Question

A 75.0 kg stunt man jumps from a balcony and falls 24.0

Answer 1

Answer:

The force that acts on the man equals 15354.75 newtons.

Explanation:

After falling through a distance of 24.0 meters the speed of the stunt man upon hitting the mattress can be obtained using third equation of kinematics as

$v^{2}=u^{2}+2gh\\\\\therefore v=\sqrt{2gh}\\\\v=\sqrt{2\times 9.81\times 24}\\\\v=21.7m/s$ ( u=0 since the man falls from rest)

Now since the man is decelerated through a distance of 1.15 meters thus the acceleration produced can be obtained from third equation of kinematics as

$v^{2}=u^{2}+2as\\\\0=u^{2}=2as\\\\a=\frac{-u^{2}}{2s}\\\\a=-\frac{(21.7^{2}}{2\times 1.15}=-204.73m/s^{2}$

Now by newton's second law the force that produced deceleration of the calculated magnitude is obtained as

$F=mass\times acceleration\\\\F=75.0\times -204.73=-15354.75Newtons$

The negative sign indicates that the direction of force is opposite of motion.