Question

A 75.0 kg stunt man jumps from a balcony and falls 24.0 m before colliding with a pile of mattresses. If the mattresses are compressed 1.05 m before he is brought to rest, what is the average force exerted by the mattresses on the stuntman?

Answer 1

# Falling 24 m from rest to reach a final velocity

using equation of motion 3, we have

v² = v₀² + 2 a y

v² = (0 m/s)² + 2 (9.8 m/s²) (24 m)

v = 470.4 m²/s²

v = 21.6 m/s

# Stopping in 1.05 m to a final velocity of 0 m/s

again, using an above formula -

v² = v₀² + 2 a y

(0 m/s)² = (21.6 m/s)² + 2 a (1.05 m)

a = - (466.5 m²/s²) / (2.1 m)

a = - 222.1 m/s²

The average force exerted by the mattresses on stuntman which will be given as :

F_net = m a

- F_M + F_g = m a

- F_M = m a - F_g

F_M = - (75 kg) (- 222.1 m/s²) + (75 kg) (9.8 m/s²)

F_M = [(16657.5 N) + (735 N)]

F_M = 17392.5 N