Question

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

mbefore colliding with a pile of mattresses. If the mattresses
arecompressed 1.15 m before he is brought to rest, what is the
averageforce exerted by the mattresses on
thestuntman?

Answer 1

The force that acts on the man equals 15354.75 newtons.

Step-by-step 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=√(2gh)\\\\v=√(2* 9.81* 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=(-u^(2))/(2s)\\\\a=-((21.7^(2))/(2* 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* acceleration\\\\F=75.0* -204.73=-15354.75Newtons`

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