Safe landin To determine AH, we assume an average force Fa FB, where FB is the buckling force of the cardboard walls. The Appendix shows that the buckle force is related to the Young s modulus and other physical parameters bp Ing F 12g2 In this case, V=cWr for a square box. Then AH can be estimated by FB△H T △H=2C For a typical box, AH is on the order of a few centimeters Stage 3: Box is Crushed Without structural integrity, the box is crushed by the applied force. Hov ever, in crushing the box, the air inside must be pushed out Let a be the surface area of the top of the box. We make the ad hoc as- sumption that when the box buckled, it is torn such that air can escape through an area a A. Provided a N O(1), we can assume incompressible flow, since the area of opening in the box is of the same order of magnitude as the area being pushed in. By conservation of mass, we obtain the the velocity of the air moving out of the box in terms of a and the velocity of the stunt person SA= ua(aa) Using conservation of energy, we equate the change in energy of the stunt person and the air leaving the box, dos dve dua dva\ The potential energy of the air does not change significantly as it is ejected, so the energy equation simplifies to dos dvs dU The energy gain of air outside the box is due to air carrying kinetic energy out of the boX dUa ldma u2=2(padua)u2 dtSafe Landings 223 To determine ∆H, we assume an average force F ≈ FB, where FB is the buckling force of the cardboard walls. The Appendix shows that the buckling force is related to the Young’s modulus and other physical parameters by FB = Y π2 12 wτ 3 2 . (2) In this case, V = 4wτ for a square box. Then ∆H can be estimated by FB∆H = 1 2 T2 S Y V, Y π2 12 wτ 3 2 ∆H = 2wτ T2 S Y , ∆H = 24 π2 TS Y 2  τ 2 . For a typical box, ∆H is on the order of a few centimeters. Stage 3: Box is Crushed Without structural integrity, the box is crushed by the applied force. How￾ever, in crushing the box, the air inside must be pushed out. Let A be the surface area of the top of the box. We make the ad hoc as￾sumption that when the box buckled, it is torn such that air can escape through an area αA. Provided α ∼ O(1), we can assume incompressible flow, since the area of opening in the box is of the same order of magnitude as the area being pushed in. By conservation of mass, we obtain the the velocity of the air moving out of the box in terms of α and the velocity of the stunt person: −usA = ua(αA) =⇒ ua = −us/α. Using conservation of energy, we equate the change in energy of the stunt person and the air leaving the box, dUs dt + dVs dt = − dUa dt + dVa dt . The potential energy of the air does not change significantly as it is ejected, so the energy equation simplifies to dUs dt + dVs dt = −dUa dt . (3) The energy gain of air outside the box is due to air carrying kinetic energy out of the box: dUa dt = 1 2 dma dt u2 a = 1 2 (ραAua) u2 a , (4)