Parachute Jumping,Falling,and LandingDavid C.Arney,BarbraS.Melendez,DebraSchnelleIntroductionIt is extremely important that leaders of airborne units understand the safety,medical,and operationalissues involved inparachute jumping.Mathematicalmodelingoftheflightandthelandingofeitherapersonoracargopackagecanhelpwiththisunderstanding.Inthissection,wedescribe,analyze,andmodelseveral components ofparachute jumping.Thesearebasicmodels usingsimplified data in orderto begin the study ofthese issues and should notbe usedforactualairborneoperations.TheArmyhasnumeroustechnicalmanualsthatgovernthetechnicalissues associatedwith airborne operations.TheSituationAnairborne jumper,whosemass is 103kg,leaps withavelocityof0.555m/sstraightout(directionofthex-axis) from the side of an airplane which is moving at avelocityof 115 m/s (direction of the y-axis).Theairplaneisflyingatanaltitude(height onthez-axis)of4000 meters.When the jumperleaves the aircraft, heishighenoughinaltitudethattheairresistancecanbeconsidered negligible,and he falls as a freelyfallingbodyfor 11.5seconds until hepulls theparachutecord.Afterthatpoint,thefall becomesathreedimensionalprojectilemotionproblemwithnon-constantaccelerationalongeachaxisasairresistancemust now be taken into account.The reference axesareestablishedasfollows(x-axisindirectionofjumperleavingplane,y-axis in direction of planewhen thejumper exits, z-axis is vertical, and the origin is on theground directlyundertheplane themoment the jumperexits).Whentheparachutecord ispulled,weassumethattheparachutedeploysimmediately and exerts a combined drag and airresistanceforce ofmagnitudeD = dv2,where d is the drag coefficient due to the parachute and air resistance,and vis thevelocityofthe jumper.Inthis case,we used=20,butnormallythevalueofdisdifficulttodetermine,sinceitdependsonsomanyvariables.Assumethedrag force isexerted only along the vertical direction(z-axis),and that theeffectof air resistance along they-direction canbemodeled as a forceofmagnitudeF=-bv, where the minus sign indicates that the force opposes the1Formerly inDepartment of Physics,USMA61
61 Parachute Jumping, Falling, and Landing David C. Arney, Barbra S. Melendez, Debra Schnelle 1 Introduction It is extremely important that leaders of airborne units understand the safety, medical, and operational issues involved in parachute jumping. Mathematical modeling of the flight and the landing of either a person or a cargo package can help with this understanding. In this section, we describe, analyze, and model several components of parachute jumping. These are basic models using simplified data in order to begin the study of these issues and should not be used for actual airborne operations. The Army has numerous technical manuals that govern the technical issues associated with airborne operations. The Situation An airborne jumper, whose mass is 103 kg, leaps with a velocity of 0.555 m/s straight out (direction of the xaxis) from the side of an airplane which is moving at a velocity of 115 m/s (direction of the y-axis). The airplane is flying at an altitude (height on the z-axis) of 4000 meters. When the jumper leaves the aircraft, he is high enough in altitude that the air resistance can be considered negligible, and he falls as a freely falling body for 11.5 seconds until he pulls the parachute cord. After that point, the fall becomes a three dimensional projectile motion problem with nonconstant acceleration along each axis as air resistance must now be taken into account. The reference axes are established as follows (x-axis in direction of jumper leaving plane, y-axis in direction of plane when the jumper exits, z-axis is vertical, and the origin is on the ground directly under the plane the moment the jumper exits). When the parachute cord is pulled, we assume that the parachute deploys immediately and exerts a combined drag and air resistance force of magnitude D = dv2, where d is the drag coefficient due to the parachute and air resistance, and v is the velocity of the jumper. In this case, we use d = 20, but normally the value of d is difficult to determine, since it depends on so many variables. Assume the drag force is exerted only along the vertical direction (z-axis), and that the effect of air resistance along the y-direction can be modeled as a force of magnitude F = -bv, where the minus sign indicates that the force opposes the 1 Formerly in Department of Physics, USMA
direction ofmotion and b is the drag coefficient duetoairresistance.Inthis case,weuseb=10.Simultaneous to theopening of the parachute,the jumper experiences a crosswind in the positive x-direction of 1.2m/s.Theforceduetothis crosswindcanbemodeledas G=b(w-v),wherebisthedragcoefficientduetoairresistance(bissetto10above),visthespeed ofthe jumper,andwisthecrosswind speed.Thecollisionbetweenthejumperandthegroundcan alsobemodeled usingmathematicsandphysics.Thereareseveral medicalandsafetyfactorsthatneedtobeconsideredforsuchacollision.Rapid deceleration of the head canbe very seriousdue tothe sheer strain on thebrain stem (shearmeansthatthe shape ofan objecthas changed asthe result offorces).Ameasureofthisshearstraintotheheadistheseverityindexl,which isdeterminedbytheequation/ = [2(v) / (g △t)j2.5 (△t),wherevisthevelocityatimpact,△tisthedurationof impact,andgistheacceleration due togravity.When the severityindexfora collision is above1000,the collision is fatal. When the value of the severity index is approximately 400,unconsciousnessandmildconcussionaretheresult.The stress on the long bones of the legs during the collision is compressive.Whenthe compression force per unit area exceeds the ultimatetensile strength of thebone(givenbelowforvariousbones,withtheirassociatedcross-sectionalareas)thebonebreaksincompression.BoneUltimateTensileAverageStrengthCross-Sectional(N/m")Area (m")Femur1.21 x 10°5.81x101.40 ×1083.23 ×10Tibia2.20 x 1084.42 x 104Spinal Cord(back)1.80 x10Spinal Cord4.42x10(neck)Table162
62 direction of motion and b is the drag coefficient due to air resistance. In this case, we use b = 10. Simultaneous to the opening of the parachute, the jumper experiences a cross wind in the positive x-direction of 1.2 m/s. The force due to this cross wind can be modeled as G = b(w - v), where b is the drag coefficient due to air resistance (b is set to 10 above), v is the speed of the jumper, and w is the cross wind speed. The collision between the jumper and the ground can also be modeled using mathematics and physics. There are several medical and safety factors that need to be considered for such a collision. Rapid deceleration of the head can be very serious due to the sheer strain on the brain stem (shear means that the shape of an object has changed as the result of forces). A measure of this shear strain to the head is the severity index I , which is determined by the equation I = [2(v) / (g Dt)]2.5 ( Dt), where v is the velocity at impact, Dt is the duration of impact, and g is the acceleration due to gravity. When the severity index for a collision is above 1000, the collision is fatal. When the value of the severity index is approximately 400, unconsciousness and mild concussion are the result. The stress on the long bones of the legs during the collision is compressive. When the compression force per unit area exceeds the ultimate tensile strength of the bone (given below for various bones, with their associated cross-sectional areas), the bone breaks in compression. Bone Ultimate Tensile Average Strength Cross-Sectional (N/m2 ) Area (m2 ) Femur 1.21 x 108 5.81 x 10-4 Tibia 1.40 x 108 3.23 x 10-4 Spinal Cord (back) 2.20 x 108 4.42 x 10-4 Spinal Cord (neck) 1.80 x 108 4.42 x 10-4 Table 1
ThecompressiveforceFexperiencedasaresultoftheimpactontheearthcanbedeterminedfroma physicsmodelusingmomentum.Momentum p isdefinedas p= mv and the Force model is F= pl t, where p is the change inmomentum and △tis the time duration of the collision.If the jumperlands ontheground,weassumethat heperformsaperfect"parachutelandingfall,"ashewastaughtinhisfreefall class.Thetimeofhisimpactwilldependuponhowsoftthegroundiswherehelands.Snoworsoftsand,forexample,wouldsignificantlyextend hisdurationofimpactandthusreducetheforceofimpact.MoreSimplifyingAssumptionsForourdiscussionandthe initial modeling ofthis situation,weusethefollowingassumptionsinadditionto oramplifyingthosealreadydiscussed:We are not considering the complexities of landing such as landing on unevenground or in the trees or in an inappropriate manner.We assume theparachute opens instantaneously and thedragforceD=dv2 takes effect immediately with no period of transition.The jumperdoes notbendortwistupon landing,and thusthebreakingtorquesof various bones and ligaments are notbeing considered.The jumperleaves the airplaneby jumpingstraightout with norotation.Theinformationgivenfortheseverityindexandthebonecompressionvaluesforthetibiaandfemurarecorrect (althoughinrealitytheyareapproximate).Eventhoughtheforcesneededtoprovidethedecelerationofthebodyproducegreattensile stresses on ligaments andtendonsand thefailurelevelforthesearemuchlowerthanforbone,weignoretheseeffectsandconcentrateonthebonesTheimpactforce istransmittedtothebones without attenuationthroughtheligamentsormusclesTime and Path of the FallThe jumper waits 11.5seconds beforepulling hisparachuteopen.Duringthisperiodoftime,heisin free fall (no air resistance).His initial velocityinthe vertical direction(z-axis)is zero,and theacceleration isdueonlytothegravitational forceof-9.80m/s2.Thus,the jumper'svertical positionat the time he deploys his parachute is given byz (t) = zo + Vzot + 0.5 azt 2. Substitution producesz(11.5)=4000+0.5 (-9.80)(11.5)2=3351.975m.His initial velocity in the X-direction is 0.555m/s,and there is no acceleration in this direction (airresistance is negligible during the first 11.563
63 The compressive force F experienced as a result of the impact on the earth can be determined from a physics model using momentum. Momentum p is defined as p = mv and the Force model is F = Dp/ Dt, where Dp is the change in momentum and Dt is the time duration of the collision. If the jumper lands on the ground, we assume that he performs a perfect "parachute landing fall," as he was taught in his free fall class. The time of his impact will depend upon how soft the ground is where he lands. Snow or soft sand, for example, would significantly extend his duration of impact and thus reduce the force of impact. More Simplifying Assumptions For our discussion and the initial modeling of this situation, we use the following assumptions in addition to or amplifying those already discussed: · We are not considering the complexities of landing such as landing on uneven ground or in the trees or in an inappropriate manner. · We assume the parachute opens instantaneously and the drag force D = dv2 takes effect immediately with no period of transition. · The jumper does not bend or twist upon landing, and thus the breaking torques of various bones and ligaments are not being considered. · The jumper leaves the airplane by jumping straight out with no rotation. · The information given for the severity index and the bone compression values for the tibia and femur are correct (although in reality they are approximate). · Even though the forces needed to provide the deceleration of the body produce great tensile stresses on ligaments and tendons and the failure level for these are much lower than for bone, we ignore these effects and concentrate on the bones. · The impact force is transmitted to the bones without attenuation through the ligaments or muscles. Time and Path of the Fall The jumper waits 11.5 seconds before pulling his parachute open. During this period of time, he is in free fall (no air resistance). His initial velocity in the vertical direction (z-axis) is zero, and the acceleration is due only to the gravitational force of -9.80 m/s2. Thus, the jumper's vertical position at the time he deploys his parachute is given by z (t) = z0 + vz0t + 0.5 az t 2 . Substitution produces z(11.5) = 4000 + 0.5 (-9.80) (11.5)2 = 3351.975 m. His initial velocity in the x-direction is 0.555 m/s, and there is no acceleration in this direction (air resistance is negligible during the first 11.5
seconds).So, his position in thex-direction isgiven by.x(t) =Xo+vxot+0.5axt2.Substitutionproducesx(11.5)=0+0.555(11.5)+0=6.3825m.Similarly,the initial velocity in the y-direction is 115m/s,and the acceleration inthis direction is zero. Thus, the position equation in the y-direction is y=yo + Vyot +0.5 ayt 2. Substitution produces y(11.5) = 0 + 115(11.5) + 0 = 1322.5 m.Therefore,thepositioncoordinateatwhichthejumperopenedhisparachute is(x,y,z)= (6.3825,1322.5,3351.975)m.When the jumper pulls the parachute rip-cord, his velocity in the x- and y-directionswill bethe same as the initial values for those velocitieswhen he leaped fromtheaircraft. His vertical velocity can be determined by the model Vz= Vzo + azt.Substitution produces Vz= 0 + (-9.8)(11.5)=-112.7 m/s.Therefore,thevelocityof thejumper whenhepullstheparachuterip-cordisgivenby (Vx, Vy, Vz) = (0.555, 115, -112.7) m/s.Thereare severalwaystomodel thefirstfewsecondsof theparachutedeployment.First, we have assumed that the chute deployment is instantaneous.Ingeneral, the jumperis falling veryfast duringthefinal phaseoffreefall andslows considerably toa constant (orterminal)velocity over some interval of timeFor our situation, we will assume, based on an experiment, that this transitionphaselasts3secondsandthatduringthistransition,the jumperfalls53metersdownward.Nowwedeterminetheterminal (constant)velocityofthe jumperwhentheparachute is deployed.Hereaches terminal velocity when the drag force oftheparachute, D = dv2, is equal to his weight, mg, so that the resultant acceleration iszero. In other words,when dy2 +mg =0 the terminal (constant) velocityisreached.Solvingforv,wegetv=-mg/d.Substitutionproduces,v=-105(-9.8)/20=7.1042m/s in thedownwarddirection.Since at terminal velocity the jumper is falling downward at a constant rate of7.1042 m/s, the time required to fall the remaining distance of 3298 meters is:(3298)/(7.1042)=464.2seconds.Thetotaltimeofthefalliscalculatedbysumming thetimespent in freefall motion andthe timewiththeparachuteopen,both beforeand afterterminal velocity was reached.Then thetotaltime equals11.5+3.0+464.2=478.7seconds.64
64 seconds). So, his position in the x-direction is given by x(t) = x0 + vx0t + 0.5 ax t 2 . Substitution produces x(11.5) = 0 + 0.555(11.5) + 0 = 6.3825 m. Similarly, the initial velocity in the y-direction is 115 m/s, and the acceleration in this direction is zero. Thus, the position equation in the y-direction is y = y0 + vy0t + 0.5 ay t 2. Substitution produces y(11.5) = 0 + 115(11.5) + 0 = 1322.5 m. Therefore, the position coordinate at which the jumper opened his parachute is (x, y, z) = (6.3825, 1322.5, 3351.975) m. When the jumper pulls the parachute rip-cord, his velocity in the x- and y-directions will be the same as the initial values for those velocities when he leaped from the aircraft. His vertical velocity can be determined by the model vz = vz0 + az t. Substitution produces vz = 0 + (-9.8)(11.5) = -112.7 m/s. Therefore, the velocity of the jumper when he pulls the parachute rip-cord is given by (vx , vy , vz ) = (0.555, 115, -112.7) m/s. There are several ways to model the first few seconds of the parachute deployment. First, we have assumed that the chute deployment is instantaneous. In general, the jumper is falling very fast during the final phase of free fall and slows considerably to a constant (or terminal) velocity over some interval of time. For our situation, we will assume, based on an experiment, that this transition phase lasts 3 seconds and that during this transition, the jumper falls 53 meters downward. Now we determine the terminal (constant) velocity of the jumper when the parachute is deployed. He reaches terminal velocity when the drag force of the parachute, D = dv2, is equal to his weight, mg, so that the resultant acceleration is zero. In other words, when dv2 + mg = 0 the terminal (constant) velocity is reached. Solving for v, we get v = - mg / d . Substitution produces, v = - 105(- 9.8)/ 20 = 7.1042 m/s in the downward direction. Since at terminal velocity the jumper is falling downward at a constant rate of 7.1042 m/s, the time required to fall the remaining distance of 3298 meters is: (3298)/(7.1042) = 464.2 seconds. The total time of the fall is calculated by summing the time spent in free fall motion and the time with the parachute open, both before and after terminal velocity was reached. Then the total time equals 11.5 + 3.0 + 464.2 = 478.7 seconds
Determining the Landing CoordinatesApplyingNewton'sSecondLawtothe jumper'smotionaftertheparachutedeploysinboththexand ydirections,weobtainthefollowingmodelsforthe equations ofmotion.x-direction:Fx=maxwhereF= b(w-v)b(w-v) = maxso,bw-bv=mama+bv=bwmx"+ bx'=bw103x"+10x=12.y-direction: Fv=maywhereF =-bv-bv=mayso,ma+bv=0my" + by = 0103y"+10y =0.Where m is the mass of the jumper, b is the coefficient of air resistance, and w isthewind speed. We solvetheseequations to obtain the position function. Theinitial conditions aretheposition andthe velocity values atthe timetheparachutewas opened.y-direction: Use the characteristic equation mr2 + br = 0, with roots r = 0and r= -(b/m). So, the form of the general solution is y(t) = Cye0 + C2e-(bt/m),where constants C and C2 are found by using the initial conditions.These conditions are at timet=O;the time at whichtheparachute is opened andairresistanceisnolongernegligible.Soatthistime,y=1322.5mandvy=115m/s (from above). Substitution produces: 1322.5 = C1 + C2and 115 =-(b/m) C2C2=-1184.5andC1=whereb=10kg/mandm=103kg.Therefore,2507. The solution is wrtten as y(t) = 2507- 1184.5 e-(10/103)t.X-direction:Inthe x-direction,the differential equation isnon-homogeneous,sotheassociatedhomogeneousequationmx"+bx=0 shouldbesolved first,and thenthenon-homogeneous equation mx"+bx =bw is solved.The characteristic equation for the homogeneous differential equation is mr2+br= 0 where the roots are: r=0 and r=-(b/m). So,the complementary solutionisXc = Ct e0 + C2 e-(bt/m). Now we solve for the particular solution, Xp, of the non-65
65 Determining the Landing Coordinates Applying Newton's Second Law to the jumper's motion after the parachute deploys in both the x and y directions, we obtain the following models for the equations of motion. x-direction: Fx = max where F = b(w - v) so, b(w-v) = max bw - bv = ma ma + bv = bw mx" + bx' = bw 103x" + 10x' = 12. y-direction: Fy = may where F = -bv so, - bv = may ma + bv = 0 my" + by' = 0 103 y" + 10y' = 0 . Where m is the mass of the jumper, b is the coefficient of air resistance, and w is the wind speed. We solve these equations to obtain the position function. The initial conditions are the position and the velocity values at the time the parachute was opened. y-direction: Use the characteristic equation mr 2 + br = 0, with roots r = 0 and r = -(b/m). So, the form of the general solution is y(t) = C1e0 + C2e-(bt/m) , where constants C1 and C2 are found by using the initial conditions. These conditions are at time t = 0; the time at which the parachute is opened and air resistance is no longer negligible. So at this time, y = 1322.5 m and vy = 115 m/s (from above). Substitution produces: 1322.5 = C1 + C2 and 115 = -(b/m) C2 , where b = 10 kg/m and m = 103 kg. Therefore, C2 = - 1184.5 and C1 = 2507. The solution is written as y(t) = 2507 - 1184.5 e-(10/103)t . x-direction: In the x-direction, the differential equation is nonhomogeneous, so the associated homogeneous equation mx" + bx' = 0 should be solved first, and then the non-homogeneous equation mx" + bx' = bw is solved. The characteristic equation for the homogeneous differential equation is mr 2 + br = 0 where the roots are: r = 0 and r = -(b/m). So, the complementary solution is: xc = C1 e0 + C2 e-(bt/m) . Now we solve for the particular solution, xp , of the non-
homogeneous equation mx" + bx' = bw. We conjecture xp = At (the conjecture of xp= A does not work). Then, Xp'= A, and xp" = 0. Substituting this solution into thedifferential equation yields m(O) + bA = bw. Therefore, A= wand Xp= wt.The general solution is x = Xc + xp Therefore, for our non-homogeneousdifferential equation, the solution is x= C, + C2 e-(bt/m) + wt , and substitution ofour parameter values produces x(t) = C1 + C2 e-(10/103)t + 1.2 t.Weuse the initial conditiontodeterminethe constants CyandC2:Notethatagainthese are the conditions at time t = O, that is,the time at which the parachuteopenedand airresistance isnolongernegligible.So,x(0)=6.3825mandvx(0)=0.555m/s (fromabove).This substitution produces 6.3825=C1 +C2and 0.555 =-(b/m)C2+w.So,C2=6.6435and C1=-0.261. We obtain the solutionx(t) = - 0.261 + 6.6435 e-(10/103)t + 1.2 t.In order to find the position coordinates ofthe jumper's landing site, we use thepositionequationsandthetotaltimeofthejump.Giventhepositionequationsfoundabove,simplysubstituteintotheseequationsthevalueofthetimewhenthejumperlands(thetotaltimeexperiencedbythe man-parachute system),t =467.2seconds.Wefind the landing position is:X=560.4m,y=2507m,andz=0m.The LandingWe needto determinethe jumper's final velocity(i.e.,hisvelocityimmediatelybeforeimpact)andtheforce ofthecollisionuponthe jumper.Thevelocityequations willbethederivatives,with respecttotime,oftheposition equations.Therefore, we computey(t) = 2507 - 1184.5 e-(10/103)ty-direction: position equation --->vy(t) = (- 1184.5)(-10/103) e-(10/103)tvelocity equation --->=115 e-(10/103)t66
66 homogeneous equation mx" + bx' = bw. We conjecture xp = At (the conjecture of xp = A does not work). Then, xp' = A, and xp " = 0. Substituting this solution into the differential equation yields m(0) + bA = bw. Therefore, A = w and xp = wt. The general solution is x = xc + xp Therefore, for our non-homogeneous differential equation, the solution is x= C1 + C2 e-(bt/m) + wt , and substitution of our parameter values produces x(t) = C1 + C2 e-(10/103)t + 1.2 t. We use the initial condition to determine the constants C1 and C2 . Note that again these are the conditions at time t = 0, that is, the time at which the parachute opened and air resistance is no longer negligible. So, x(0) = 6.3825 m and vx (0)= 0.555 m/s (from above). This substitution produces 6.3825 = C1 + C2 and 0.555 = - (b/m) C2 + w. So, C2 = 6.6435 and C1 = - 0.261. We obtain the solution x(t) = - 0.261 + 6.6435 e-(10/103)t + 1.2 t. In order to find the position coordinates of the jumper's landing site, we use the position equations and the total time of the jump. Given the position equations found above, simply substitute into these equations the value of the time when the jumper lands (the total time experienced by the man-parachute system), t = 467.2 seconds. We find the landing position is: x = 560.4 m, y = 2507 m, and z = 0 m. The Landing We need to determine the jumper's final velocity (i.e., his velocity immediately before impact) and the force of the collision upon the jumper. The velocity equations will be the derivatives, with respect to time, of the position equations. Therefore, we compute y-direction: position equation -> y(t) = 2507 - 1184.5 e-(10/103)t velocity equation -> vy (t) = (- 1184.5)(-10/103) e-(10/103)t = 115 e-(10/103)t
x(t) = - 0.261 + 6.6435 e-(10/103)t + 1.2 tx-direction: position equation --->vx(t) = (6.6435)(-10/103) e-(10/103)t + 1.2velocity equation--->= -0.645 e-(10/103)t + 1.2At the time of impact, the jumper has reached terminal velocity in the z-direction.Therefore, we know vz(t) =-7.1042 m/s. Since we determined the total time of thefall is 478.7 seconds, and he opened his parachute at 11.5 seconds, we areinterested in his velocity at 467.2seconds.This is the time value we substitute intothevelocity equations aboveto obtain the velocity upon impact in each direction?x(467.2) = -0.645 e-(10/103)(467.2) + 1.2 = 1.20 m/sVy(467.2) = 115 e-(10/103)(467.2) = 0.0 m/sv(467.2) = -7.1042 m/sIn vector notation, we write, v= vxi + vyj + Vz k, so v= (1.20 i + 0 j- 7.1042 k)m/s.The magnitude of the velocity vector gives the speed of the jumper at the time ofimpact.Inordertodetermine the speedweusethePythagoreantheorem:speed = / (vx +vy? + v2)= / (1.2)2 + (0)2 + (-7.1042)2 = 7.2048 m/s.WhenthejumpercollideswiththegroundheexperiencesachangeinmomentumAp, given by Pfinal - Pinitial.After the jumper collides with the ground, he has nomomentum. Therefore, Prinal = o. Additionally, if we assume that the collisiontakes place along a singledirection,wecan use the speed on impacttodeterminePinitial. Momentum is given by p=mv, where m is the mass of the jumper and vishis velocity at impact. We can determine the initial momentum as Pinitial = mv =(103)(7.2048)=742.0944kgm/s.Sincethefinal momentumis0,thechange inmomentump is0-742.0944=-742.0944kg-m/s.This can also bedescribed as742.0944N-s,whereN=1kg-m/s2 indicates force in units of Newtons.Wedeterminetheforceof impactusingtheequationF=Ap/△t.Ifweassumethe jumperlandedinthickbusheswithatimeof impactof200milliseconds(0.200seconds),we calculateF=(742.0944)/(.200)=3,710.472N.67
67 x-direction: position equation -> x(t) = - 0.261 + 6.6435 e-(10/103)t + 1.2 t velocity equation -> vx (t) = (6.6435)(-10/103) e-(10/103)t + 1.2 = -0.645 e-(10/103)t + 1.2 At the time of impact, the jumper has reached terminal velocity in the z-direction. Therefore, we know vz (t) = - 7.1042 m/s. Since we determined the total time of the fall is 478.7 seconds, and he opened his parachute at 11.5 seconds, we are interested in his velocity at 467.2 seconds. This is the time value we substitute into the velocity equations above to obtain the velocity upon impact in each direction: vx (467.2) = -0.645 e-(10/103)(467.2) + 1.2 = 1.20 m/s vy (467.2) = 115 e-(10/103)(467.2) = 0.0 m/s vz (467.2) = - 7.1042 m/s In vector notation, we write, v = v x i + vy j + vz k, so v = (1.20 i + 0 j - 7.1042 k ) m/s. The magnitude of the velocity vector gives the speed of the jumper at the time of impact. In order to determine the speed we use the Pythagorean theorem: speed = (vx 2 + vy 2 + vz 2) = (1.2)2 + (0)2 + (- 7.1042)2 = 7.2048 m/s. When the jumper collides with the ground he experiences a change in momentum Dp, given by pfinal - pinitial. After the jumper collides with the ground, he has no momentum. Therefore, pfinal = 0. Additionally, if we assume that the collision takes place along a single direction, we can use the speed on impact to determine pinitial. Momentum is given by p=mv, where m is the mass of the jumper and v is his velocity at impact. We can determine the initial momentum as pinitial = mv = (103)(7.2048) = 742.0944 kg·m/s. Since the final momentum is 0, the change in momentum Dp is 0 - 742.0944 = -742.0944 kg·m/s. This can also be described as 742.0944 N·s, where N = 1 kg·m/s2 indicates force in units of Newtons. We determine the force of impact using the equation F = Dp / Dt . If we assume the jumper landed in thick bushes with a time of impact of 200 milliseconds (0.200 seconds), we calculate F = (742.0944) / (.200) = 3,710.472 N
Injuries?Fromtheresultswehavecalculated,wecandetermineif the jumpersustainedanyinjuries. Table 1 gave us the ultimate tensile strength in N/m2 and the averagecross-sectional area inm2 for four bones vulnerable to injuryinthis type of fall. Bymultiplyingthetensilestrengthandthecrosssectionweobtainthemaximumforceeachboneisabletosustainwithoutbreaking.Table2providesthisinformation.BoneUitimate TensileAverageMaxStrengthCross-SectionalSustainable(N/m)Area (m")Force (N)Femur1.21x10°5.81x10703011.40 x10°Tibia3.23x10452202.20x10°4.42x1097240SpinalCord (back)1.80x10°4.42x10SpinalCord (neck)79560Table 2FromTable2weseethataforceof3.710.47Nwouldnotbreakanyofthefourbonesanalyzed.Finally,wecheckforthepossibilityofinjurytotheheadbyshearstrain.Weusethe severityindex,l,todeterminepossibleinjury.Wecalculate/=[2(v)/(g t)j2.5 ( t) =[2(7.2048) / [(9.8)(.200)) j2.5 (.200) = 29.310. Theseverity indexof29.310isnotclosetothe concussionlevel of400.Therefore,weconclude that there should be no serious injuries to the head caused bythislanding.ConclusionWehave used mathematical modeling in a parachute jump and landing scenario toanalyze some of thefactors associated withsuchanactivity.Certainly,moredetailed and sophisticated analyses could be performedto understand more of thecomponents of parachute jumping. There are many issues that this scenario hasignoredor that were assumed to hold that couldusefurther study.For instance,thetransitionperiodfromfreefall to steady-statefalling witha deployed chutecontainsnumerousfactorsthatareworthyof morestudy(seetheexercises).Alsoparachutedeploymentofheavyequipment likevehicles,weapon systems,andsupplies is alsoworthy of study to insure the cargo lands safely and in the properlocation.Exercises1.Arethere additional assumptions being made that shouldbe explicitly listed?What arethey?Write a shortessaythat describes therole assumptions playinthemodelingprocess.68
68 Injuries? From the results we have calculated, we can determine if the jumper sustained any injuries. Table 1 gave us the ultimate tensile strength in N/m2 and the average cross-sectional area in m2 for four bones vulnerable to injury in this type of fall. By multiplying the tensile strength and the cross section we obtain the maximum force each bone is able to sustain without breaking. Table 2 provides this information. Bone Ultimate Tensile Average Max Strength Cross-Sectional Sustainable (N/m2 ) Area (m2 ) Force (N) Femur 1.21 x 108 5.81 x 10-4 70301 Tibia 1.40 x 108 3.23 x 10-4 45220 Spinal Cord (back) 2.20 x 108 4.42 x 10-4 97240 Spinal Cord (neck) 1.80 x 108 4.42 x 10-4 79560 Table 2 From Table 2 we see that a force of 3,710.47 N would not break any of the four bones analyzed. Finally, we check for the possibility of injury to the head by shear strain. We use the severity index, I, to determine possible injury. We calculate I = [2(v)/(g Dt)]2.5 ( Dt) = [2( 7.2048) / [(9.8)(.200)] ]2.5 (.200) = 29.310. The severity index of 29.310 is not close to the concussion level of 400. Therefore, we conclude that there should be no serious injuries to the head caused by this landing. Conclusion We have used mathematical modeling in a parachute jump and landing scenario to analyze some of the factors associated with such an activity. Certainly, more detailed and sophisticated analyses could be performed to understand more of the components of parachute jumping. There are many issues that this scenario has ignored or that were assumed to hold that could use further study. For instance, the transition period from free fall to steady-state falling with a deployed chute contains numerous factors that are worthy of more study (see the exercises). Also, parachute deployment of heavy equipment like vehicles, weapon systems, and supplies is also worthy of study to insure the cargo lands safely and in the proper location. Exercises 1. Are there additional assumptions being made that should be explicitly listed? What are they? Write a short essay that describes the role assumptions play in the modeling process
2.Use the severityIndexformula todetermine the shear strain on the head duringthe3 second transition period fromfreefall to the chute-deployed steady-stateconstant velocity.3. What would happen if the jumper in the scenario landed on a very hard surface,likerocks,andtheinterval of impactwasreducedto △t=0.01seconds?Doyoupredict any injuries for the jumper?4. Solve forthelanding time and location, if the same scenario and model areusedwithanexitaltitudeof6,000meters.5.Analyze the injury situation, if the same scenario and model are used with anexitaltitudeof6,000meters.6. In this scenario the transition phasefrom free-fall tosteady-state descent with afullydeployedchuteis3secondswithaaltitudechangeof 53meters.Thevelocitychangesfrom112.7m/s to7.1m/s over that interval.Willalinearmodel fitthisdata?Why orwhynot? Find a second-order (quadratic polynomial)tofit this data.AcknowledgementsThisscenariowasdevelopedfromthefollowingarticle:Melendez,BarbraandSchnelle,Debra,“ParachutePanic",InterdisciplinaryLivelyApplicationprojects(DavidArney,editor),Washington:MathematicalAssociationof America, 1997.69
69 2. Use the severity Index formula to determine the shear strain on the head during the 3 second transition period from free fall to the chute-deployed steady-state constant velocity. 3. What would happen if the jumper in the scenario landed on a very hard surface, like rocks, and the interval of impact was reduced to Dt=0.01 seconds? Do you predict any injuries for the jumper? 4. Solve for the landing time and location, if the same scenario and model are used with an exit altitude of 6,000 meters. 5. Analyze the injury situation, if the same scenario and model are used with an exit altitude of 6,000 meters. 6. In this scenario the transition phase from free-fall to steady-state descent with a fully deployed chute is 3 seconds with a altitude change of 53 meters. The velocity changes from 112.7 m/s to 7.1 m/s over that interval. Will a linear model fit this data? Why or why not? Find a second-order (quadratic polynomial) to fit this data. Acknowledgements This scenario was developed from the following article: Melendez, Barbra and Schnelle, Debra, “Parachute Panic”, Interdisciplinary Lively Application projects (David Arney, editor), Washington: Mathematical Association of America, 1997