Longer to Rise or to Fall? Answer: Longer to fall, for Newton models with air resistance. Example 3, Edwards-Penney 2.3 ============================= A bolt is shot straight upward (along the gravity vector) with initial velocity v1(0)=49 m/s from a crossbow at ground level. Assume air resistance proportial to the square of the velocity with drag rho=0.0011, giving the equations (12) (d/dt)v1(t) = - 9.8 - 0.0011 v1(t)^2 [v1=rise] (15) (d/dt)v2(t) = - 9.8 + 0.0011 v2(t)^2 [v2=fall] Compare the rise and fall times for this Newton model with the models (1) (d/dt)v3(t) = -9.8 [v3=velocity, no air resistance] (4) (d/dt)v4(t) = - 9.8 - 0.04 v4(t) [V4=velocity, linear air resistance] Results ======= Model Max Flight Rise Fall Impact used Height time time time velocity ===== ====== ====== ==== ==== ======== v3 122.5 10.00 5.00 5.00 49.00 v4 108.28 9.41 4.56 4.85 43.23 v1 & v2 108.47 9.41 4.61 4.80 43.49