Question

A person jumps from the roof of a house 4.0 m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 m. The mass of his torso (excluding legs) is 45 kg.

(a) Find his velocity just before his feet strike the ground.
m/s (downward)
(b) Find the average force exerted on his torso by his legs during deceleration.
Magnitude
N Direction

Answer 1

first of all, the time he takes to fall is ~0.90 seconds. For (a), simply multiply 0.9 seconds by 9.8 and ka-ching you have an answer of 8.82 meters per second.
For (b), we have to remember that f=ma, or force = mass * acceleration. For this, we just take the velocity (our acceleration in this case) which is 8.82, and multiply it by 45 kg, to get 396.9 newtons of force. BUT WAIT he's decelerating over 0.7 meters! This means that you must divide by two and get 198.45 newtons average. I believe this is right, because the force exerted on his torso by his legs would be the same as the force exerted on his legs by his torso because of newtons law of the equal and opposite reaction, but I would check with your textbook's answer key to be sure

Answer 2

I didnt take into account the 0.7 but I dont think that matters...

Answer 3

oh yea the way you figure out the time it takes to fall a certain distance is (using seconds for t and meters for d)\[\sqrt{\frac{ d }{ 4.9 }}=t\]

Answer 4

this is derived from 4.9t^2=d, which is the distance covered (on earth, in meters) over a certain amount of time (in seconds)