Answer:
the maximum height of Joe's ball will be 4 times Bill's ball.
Explanation:
let bill's velocity be v then Joe's velocity is 2v.
and initial velocities of both bill and Joe are 0
for Bill
[tex]v^2-u^2= 2gh[/tex]
[tex]h=\frac{v^2}{2gh}[/tex]
for Joe
[tex](2v)^2-u^2= 2gh'[/tex]
[tex]h'=\frac{4v^2}{2g}[/tex]
thus we can write that
h'=4h
the maximum height of Joe's ball will be 4 times Bill's ball.
