Asked by Emily Grossman

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?

Answers

Answered by drwls
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.
Answered by telly
.307
Answered by Anonymous
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
Answered by Anonymous
A 57kg trampoline artist jumps vertically upward from the top of a platform with a speed of 4.7m/s. How fast is he going as he lands on the trampoline 3.8m below?
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