When jumping straight down, you can be seriously injured if you land stiff-legge

Question

When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 64.0-kg man just before contact with the ground has a speed of 7.43 m/s. (a) In a stiff-legged landing he comes to a halt in 4.16 ms. Find the magnitude of the average net force that acts on him during this time. (b) When he bends his knees, he comes to a halt in 0.295 s. Find the magnitude of the average net force now. (c) During the landing, the force of the ground on the man points upward, while the force due to gravity points downward. The average net force acting on the man includes both of these forces. Taking into account the directions of the forces, find the magnitude of the force applied by the ground on the man in part (b).

Explanation / Answer

F=ma we have m we need a To find a, we can use one of our beloved kinematic equations (yes, we aren't through with them; they show up all semester, so don't forget them) 7.43 m/s = Vo 0.00416s = t 0m/s = Vf a = ? do we have an equation that doesn't need more than we have and provides us what we want? Yes, Vf = Vo + at a = -Vo/t = (7.43m/s)/(0.00416s) = 1786 m/s^2 (this doesn't include his weight yet, see part c) now use this in F = ma = 64kg*1786m/s^2 = 114,000 N (a) answer He feels an extra force of 114,000N (b) a = -Vo/t = (7.43m/s)/(0.295s) = 25.19 m/s^2 (this doesn't include his weight yet, see part c) now use this in F = ma = 64kg*25.19m/s^2 = 1,612 N (b) answer He feels an extra force of 1,612 N (c) the answers to a and b don't include his normal weight yet. We want to add mg to the force that he feels. (Both forces push your feet and knees together, so we add them) mg = 64.0kg*9.8m/s^2 = 627 N So, the magnitude of the force applied by the ground on the man is 1,612N + 627N = 2239N or with three significant figures 2,240 N

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