Answer:
The average force exerted on the man by the ground therefore is 153.319.53 N
Explanation:
Given the following information
Mass of man, m = 75 kg
height of fall, h = 0.48 cm
velocity just before landing, v = 4.43 m/s
We therefore have
The work required to break the fall is equal to the kinetic energy of motion, just before touching the ground
Work done = Energy to absorb Kinetic Energy KE = 0.5·m·v²= F·h
Where:
F = Force required to break the fall
Therefore the force, F = (0.5·m·v² )/h
= 0.5×75 kg ×(4.43 m/s)²/(0.0048 m) = 153319.53 N
The average force exerted on him by the ground is therefore
= 153319.53 N.
Answer:
23 m
Explanation:
If acceleration is uniform then
distance = ( average velocity ) × time
= ( (initial velocity +final velocity )/2 ) × time
= (( 1.5 + 1.0) /2 ) × 18.4
= ( 2.5/2 ) × 18.4
= 1.25× 18.4
= 23 m
I'm not entirely sure, but I think the first is A, and the second is inverted.
