physics question?

Question

A person with mass of 25 kg on a trampoline jumps up and lands. If the person's height was 3 meters and after landing on the trampoline he continues down for another 1 meter before coming to a stop. What was the force exerted by the trampoline on the person?

Answer 1

For any object the ratio of distances (free falling from hight h, to stopping over distance s) gives the force in g-s.

Thus the trampoline jumper experiences a force of 3g, or a force of 3 times his/hers weight (weight not mass, and this is an important distinction). So ig you take g=9.81m/s^2 then the force exerted by the trampoline is
3 x 25 x 9.81 = 735.75 Newtons

Answer 2

The force exerted by the trampoline on the person at rest is the weight of that person
W=F=mg (Big deal :-) ) When a trampoline ‘returns the favor’ it is a bit different.

When the person hit the trampoline coming down from a 3m height in has a kinetic energy equal to potential energy of
P=mgh
The trampoline absorbs the energy by allowing the person go down 1 meter. We cam model a trampoline as a spring

F=kx where
k- coefficient
x – displacement

The energy in the spring is E=.5kx^2

Since the energy of the body was converted to the spring (ospring of th etrampoline) we have

P=E
mgh=.5kx^2
then k=2mgh/x^2

and F=kx=2mgh/x

so F=2 (25kg) 9.81 m/s^2 (3m)/(1m)= 1471.5N

I hope it helps.