Answer:4.04 m
Step-by-step explanation:
mass of athlete
![m=57 kg](https://img.qammunity.org/2020/formulas/physics/high-school/5jx83s1mfly7z6qk3pdy7igp74vi72wcuw.png)
initial speed
![u=8.9 m/s](https://img.qammunity.org/2020/formulas/physics/high-school/pu5bnm9xbt700j4a3p30ffidk897r30p83.png)
To get the maximum height h of Athlete we conserve energy i.e.
Kinetic Energy of Athlete=Potential energy gained by athlete
![(1)/(2)* mu^2=m\cdot g\cdot h](https://img.qammunity.org/2020/formulas/physics/high-school/wqi0ywya1h152qhqqe3pp6mmz4xhej96r9.png)
![h=(v^2)/(2g)](https://img.qammunity.org/2020/formulas/physics/high-school/4bfxxx07vjk0kj73siw9pv542b084g51kt.png)
![h=(8.9^2)/(2* 9.8)](https://img.qammunity.org/2020/formulas/physics/high-school/zfpu9uj1lhv3p18pwqho789k5ka3x62qz2.png)
![h=4.04 m](https://img.qammunity.org/2020/formulas/physics/high-school/zsieavkj813e8mrnkjin5ws2agqsia0i29.png)
(b)Speed at half of maximum height
Considering v be the velocity at half of maximum height
conserving Energy we can write
![(mu^2)/(2)=(mv^2)/(2)+mg(h)/(2)](https://img.qammunity.org/2020/formulas/physics/high-school/cz8e0w3k04uk90lmc9w0d1qslxpsd5da0x.png)
(as
)
thus
![v^2=(u^2)/(2)](https://img.qammunity.org/2020/formulas/physics/high-school/rxsq4tm96uzxe4djp3mljpb5a4jwc0ng9v.png)
![v=(u)/(√(2))](https://img.qammunity.org/2020/formulas/physics/high-school/y8uvg7n16g6dc9cokqpdu4vy1jmio5vvkj.png)
![v=(8.9)/(√(2))](https://img.qammunity.org/2020/formulas/physics/high-school/mmvptza670jv4eq6rtqo7iegq9qph7p2eb.png)
![v=6.29 m/s](https://img.qammunity.org/2020/formulas/physics/high-school/ko7k7laf47gh3pu3s90a1nioq9yfjmy81l.png)
So Athlete interact with the gravitational Field of Earth