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 75 kg man just before contact with the ground has a speed of 7.0 m/s.
(a) In a stiff-legged landing he comes to a halt in 1.6 ms. Find the average net force that acts on him during this time.
(b) When he bends his knees, he comes to a halt in 0.12 s. Find the average 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 these forces, find the force of the ground on the man in parts (a) and (b).
(i) stiff legged landing
(ii) bent legged landing
(a) Remember that his momentum befor impact, 75x7 = 525 kg m/s, equals the product of the average force and the time that it acts. Therefore
(avg. force) = (525 kg m/s)/1.6*10^-3 s
The answer will be in Newtons.
(b) Use the same formula with the longer time interval.
(c) (avg. Force on ground - M g) = Avg force on Man during impact.
Therefore subtract Mg = 735 N from the answers of (a) and (b)
Thanks you so much for your help. In this question I got for
a. 328125 N
b. 4375 N
c. stiff legged landing
327390 N
bent legged landing
4375 N - (75kg x 9.8)= 3640 N
All of the answers are correct except the bent legged landing in part c. Please tell me where I went wrong.
Your last answer seems right to me
the solution to your problem is
4375 N - (75kg x -9.8) = 5110 N
2 answers