Question

A 58.5-kg athlete leaps straight up into the air from a trampoline with an initial speed of 8.8 m/s. the goal of this problem is to find the maximum height she attains and her speed at half maximum height.

Answer 1

1) Maximum height
This is a uniformly accelerated motion with initial speed
`v_0 = 8.8 m/s` and acceleration
`a=g=-9.81 m/s^2` (the negative sign is required because the acceleration is directed downward, in the opposite direction of the motion, which is directed upward). The distance covered (S) in a uniformly accelerated motion is related to the acceleration and the initial and final velocity by the relationship

`2aS=v_f^2-v_i^2`
At maximum height, S=h and the final velocity is zero:
`v_f =0`. So we can solve for h and find the maximum height:

`2ah=-v_i^2`

`h= (-v_i^2)/(g)= (-(8.8 m/s)^2)/(-9.81 m/s^2)=3.95 m`

2) Speed at half maximum height
Half maximum height corresponds to
`(h)/(2)= (3.95 m)/(2)=1.975 m`. To find the speed at this height, we can use the same relationship used in the previous step:

`2a (h)/(2)=v_f^2-v_i^2`
where
`v_f` is the speed at half maximum height.
Re-arranging and substituting numbers, we find

`v_f = \sqrt{v_i^2+(2g (h)/(2)) }= √((8.8m/s)^2+2(-9.81m/s^2)(1.975m))=6.22m/s`