A 75.0 kg stunt man jumps from a balcony and falls 22.5 m before colliding with a pile of mattresses. If the mattresses are compressed 1.00 m before he is brought to rest, what is the average force exerted by the mattresses on the stuntman?
2006-10-15 11:15:23 · 1 answers · asked by rhondasumpter 2 in Science & Mathematics ➔ Physics
F = ma
m = 75 kg
a = acceleration which is the change in velocity over time.
The stunt man free falls 22.5 meters accelerated by gravity.
y = 1/2 g t1^2 y = 22.5 meters g = 9.8 m/s^2
t1 = sq root (2 y / (g))
Velocity at the end of the free fall = g t1
He travels 1 meter into the mattress.
y = Vt2 - 1/2 a t2^2 Where y = 1, V = gt1 a = V/t2
1 = Vt2 - 1/2 (V) t2 Solve for t2 = 1/(V-V/2)
a = V/t2
F = M * a Plug and chug
2006-10-15 12:09:46 · answer #1 · answered by Roadkill 6 · 0⤊ 0⤋