Question

A gymnast spring vertical upward from a trampoline as in figure. The gymnast leaves the trampoline at height of 1.20m and reaches a maximum height of 4.80m before falling back down. all heights are all measured with respect to the ground. Ignoring air resistance, determine the initial speed “vo” with witch the gymnast leaves the trampoline.

Answer 1

Answer:

The initial velocity of the gymnast is 8.5 m/s.

Explanation:

We can use the kinematic equation

$v_f^2= v_o^2+2ad$

to figure out the initial velocity $v_o$ of the gymnast.

Now, when the gymnast reaches the maximum height,  the distance he has traveled is $d = 4.8m- 1.2m = 3.6 m$, and his velocity is zero; therefore $v_f =0$.

Thus, we have

$0 = v_0^2+2(-10m/s^2)(3.6m)$

$v_0^2=72m^2/s^2$

$\boxed{v_0= 8.5m/s}$