Question

A 62 kg rock climber falls off the wall at the gym. Immediately before she hits the padded floor, her velocity is 7.9 m/s, 15° away from the vertical. She comes to a stop 0.75 s later. What vertical force did the padded floor exert on the rock climber? If the floor was not padded and she instead came to a stop in 0.25 seconds, what vertical force would the floor have exerted on her?

Answer 1

Answer:

Explanation:

Given

mass of climber= 62 kg

velocity of climber before it hits=7.9 m/s at an angle of $15^{\circ}$ from vertical

therefore change in momentum in vertical direction is

$\Delta P=m(v_i-v_f)$

$\Delta P=62\left ( 7.9cos15-(7.9cos15)\right )$

$\Delta P=2\times \62\times 7.9cos15=946.22 N.s$

there is no change in horizontal momentum

and impulse=change in momentum

F.dt=946.22

$F=\frac{946.22}{0.75}=1.261 kN$

For non padded floor

$F=\frac{946.22}{0.25}=3748.88 =3.748 kN$