posted 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?
a) When hitting the trampoline H=3.0 m below, the kinetic energy will have increased
from (1/2)MVo^2 to
(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.
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
Then solve for his highest position using the time interval
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.