Question

A-72kg trampoline artist jumps vertically upward from the top of a platform with a speed of 4.5m/s.
a) How fast is he going as he lands on the trampoline 2.0m below?
b) If the trampoline behaves like a spring of spring constant 5.8x10^4N/m, how far does he depress it?
Please help

Answer 1

for part a) I use \(KE_{initial}+ PE_{initial}= KE_{final}+KE_{final}\)
\[\frac{1}{2}mv_i^2+mgh_0 = \frac{1}{2}mv_f^2+mgh_f\]
\(v_f= 7.7m/s\)
but I don't get part b which answer is -0.28 m

Answer 2

@ybarrap hey, help me, friend

Answer 3

It's asking you to use Hooke's Law. Do you know it?

Answer 4

yes, F= -kx

Answer 5

the net is soooo bad here

Answer 6

(I know, I think the server is slow today)

I think this is the approach:

Hooke's Law is F=kx, where k is a constant and x is the displacement away from rest:

Using Newton's 2nd Law:
$$
F(x)=ma(x)=-mg-(-kx)=-mg+kx
$$
At maximum displacement, \(x_{max}\), velocity is zero and not upward or downward acceleration, i.e. a=0.
So,
$$
F(x_{max})=ma(x_{max})=0=-mg+kx_{max}
$$
Now, let k=5.8x10^4N/m and solve for \(x_{max}\), the maximum displacement.

Answer 7

It says I'm typing a reply but I'm not lol

Answer 8

Anyone tried using a graphics tablet to work with maths software ?