# Physics

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A 65kg trampoline artist jumps vertically upward from the top of a platform with a speed of 5.0m/s. a) How fast is he going as he lands on the trampoline, 3.0m below. b) if the trampoline behaves like a spring with a spring stiffness constant 6.2*10^4N/m, how far does he depress it?

• Physics -

a) When hitting the trampoline H=3.0 m below, the kinetic energy will have increased
from (1/2)MVo^2 to
(1/2)MV1^2 =
(1/2)MVo^2 + MgH

Solve for V1
V1^2 = Vo^2 + 2gH

b) (1/2)MV1^2 = (1/2)kX^2
Solve for X, the trampoline displacement.

• Physics -

.307

• Physics -

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• Physics -

To solve for the final velocity you need the time and the point where his velocity was zero.

v=v0+at since your solving for a vf=0
v0/a=t
5/9.8=0.51s

Then solve for his highest position using the time interval

x=x0+v0t+1/2(at^2)
x=3+(5)(0.51)+1/2(-9.8)(0.51^2)
=4.28m

Next, solve use energy equations setting the potential energy at the top of his jump equal to the kinetic energy when he hits the trampoline.

1/2(mv^2)=mgh
1/2(v^2)=gh
v=sqrt(2gh)
v=sqrt(2(9.8)(4.28))=9.2m/s

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