v = g t
h = (1/2) g t^2
1.5 = 4.9 t^2 solve for t
then
v = 9.81 t
that is part a
.49 cm = .0049 meter
force * .0049 = (1/2) m v^2
A 72 kg man jumps fro a window 1.5 m above a sidewalk.
The acceleration of gravity is 9.81 m/s^2.
a) what is his speed just before his feet strike the pavement? Answer in units of m/s.
b) If the man jumps with his knees and ankles locked, the only cushion for his fall is approximately 0.49 cm in the pads of his feet.
Calculate the magnitude of the average force exerted on him by the ground in his situation. Answer in units of N.
1 answer
1 answer