Question

A 100-kg athlete is performing a vertical leap to test his leg strength. He leaves the ground at 3.20 m/s, zooms upward to some maximum altitude, and then returns to the ground. As he lands, he bends his legs and crouches down, so that he moves 30 cm downward as he comes to a stop. What average force did the athlete’s legs exert on the ground as he landed?

Answer 1

Answer:

2687.7 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

When he will reach the ground his speed will be 3.2 m/s but will travel 30 cm down

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.2^2}{2\times 0.3}\\\Rightarrow a=-17.067\ m/s^2$

Force

$F=m(g-a)\\\Rightarrow F=100\times (9.81-(-17.067))\\\Rightarrow F=2687.7\ N$

The force on the legs will be 2687.7 N