Question

A student (m = 57 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of 0.04 s. The average force exerted on him by the ground is +18000 N, where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.

Answer 1

y=8.14m

Step-by-step explanation:

Hi! Let's solve this!

First we have to know the speed of fall to then calculate the height.

We have the following data:

F = 18000N

m = 57kg

t = 0.04s

g=9.8m/
`s^(2)`

We use the formula

F * t = m * v

v = (F * t) / m = (18000N * 0.04s) / 57kg

v = 12.63m / s

Then we use the formula

`vf^(2) =vi^(2)+2*g*y`

`y=(vf^(2) -vi^(2) )/(2*g)`

`y=((12.63)^(2)(m/s)^(2) )/(2*9.8m/s^(2) )`

y=8.14m