Question

A 57.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 8.9 m/s. The goal of this problem is to find the maximum height she attains and her speed at half maximum height. (a) What are the interacting objects and how do they interact?

Answer 1

Answer:4.04 m

Explanation:

mass of athlete $m=57 kg$

initial speed $u=8.9 m/s$

To get the maximum height h of Athlete we conserve energy i.e.

Kinetic Energy of Athlete=Potential energy gained by athlete

$\frac{1}{2}\times mu^2=m\cdot g\cdot h$

$h=\frac{v^2}{2g}$

$h=\frac{8.9^2}{2\times 9.8}$

$h=4.04 m$

(b)Speed at half of maximum height

Considering v be the velocity at half of maximum height

conserving Energy we can write

$\frac{mu^2}{2}=\frac{mv^2}{2}+mg\frac{h}{2}$

$\frac{mu^2}{4}=\frac{mv^2}{2}$   (as $\frac{1}{2}\times mu^2=m\cdot g\cdot h$)

thus $v^2=\frac{u^2}{2}$

$v=\frac{u}{\sqrt{2}}$

$v=\frac{8.9}{\sqrt{2}}$

$v=6.29 m/s$

So Athlete interact with the gravitational Field of Earth