Example Ballistic motion Consider the free-Hlight motion of a projectile which is initially launched with a velocity vo vo cos oi to sinoj. If we neglect air resistance, the only force on the projectile is the weight,which causes the projectile to have a constant acceleration a=-g3. In component form this equation can be written as dv / dt= 0 and doy/dt Integrating and imposing initial conditions, we get o, Uy=to sin- gt where we note that the horizontal velocity is constant. A further integration yields the trajectory (uo cos o)t, y=yo+(vo sin o niCi we recog nize as the equation of a parabola The maximum height, ymh, occurs when vy(tmh)=0, which gives tmh=(vo/g) sin o, or, The range, Ir, can be obtained by setting y= yo, which gives tr=(2vo/g)sin o, or sin o cos o sin(2o) We see that if we want to maximize the range ar, for a given velocity vo, then sin(2)=l, or =450 Finally, we note that if we want to model a more realistic situation and include aerodynamic drag forces of he form, say, -ku", then we would not be able to solve for a and y independently, and this would make the problem considerably more complicated(usually requiring numerical integration) ADDITIONAL READING J. L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 2/1,2/3,2/4Example Ballistic Motion Consider the free-flight motion of a projectile which is initially launched with a velocity v0 = v0 cos φi + v0 sin φj. If we neglect air resistance, the only force on the projectile is the weight, which causes the projectile to have a constant acceleration a = −gj. In component form this equation can be written as dvx/dt = 0 and dvy/dt = −g. Integrating and imposing initial conditions, we get vx = v0 cos φ, vy = v0 sin φ − gt , where we note that the horizontal velocity is constant. A further integration yields the trajectory x = x0 + (v0 cos φ) t, y = y0 + (v0 sin φ) t − 1 2 gt2 , which we recognize as the equation of a parabola. The maximum height, ymh, occurs when vy(tmh) = 0, which gives tmh = (v0/g) sin φ, or, ymh = y0 + v 2 0 sin2 φ 2g . The range, xr, can be obtained by setting y = y0, which gives tr = (2v0/g) sin φ, or, xr = x0 + 2v 2 0 sin φ cos φ g = x0 + v 2 0 sin(2φ) g . We see that if we want to maximize the range xr, for a given velocity v0, then sin(2φ) = 1, or φ = 45o . Finally, we note that if we want to model a more realistic situation and include aerodynamic drag forces of the form, say, −κv2 , then we would not be able to solve for x and y independently, and this would make the problem considerably more complicated (usually requiring numerical integration). ADDITIONAL READING J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 2/1, 2/3, 2/4 7