# Thread: Force of ground when jumping

1. ## Force of ground when jumping

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 6.0 m/s.

(a) In a stiff-legged landing he comes to a halt in 2.0 ms. Find the average net force that acts on him during this time.
in Newtons

(b) When he bends his knees, he comes to a halt in 0.08 s. Find the average force now.
in Newtons

(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.)
for the stiff legged landing in Newtons

and

(b.)
for the bent legged landing in Newtons

most of this seems straight forward i believe, but isn't this just F = M * A ? thankyou, just need some clarification just to make sure i have the logic.

2. Originally Posted by rcmango
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 6.0 m/s.

(a) In a stiff-legged landing he comes to a halt in 2.0 ms. Find the average net force that acts on him during this time.
in Newtons

(b) When he bends his knees, he comes to a halt in 0.08 s. Find the average force now.
in Newtons

(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.)
for the stiff legged landing in Newtons

and

(b.)
for the bent legged landing in Newtons

most of this seems straight forward i believe, but isn't this just F = M * A ? thankyou, just need some clarification just to make sure i have the logic.
It employs the impulse-momentum theorem, which is slightly more general than $\displaystyle \sum F = ma$. Because the impluse uses an average force, the acceleration does not need to be constant. (For the record the most general, and most powerful, form of Newton's 2nd Law is$\displaystyle \sum F = \frac{dp}{dt}$.)

You've gotten some examples on how to use impulse-momentum. Give this one another try and respond if you are still having problems.

-Dan