A child slides without friction from a height `h` along a curved water slide (figure). She is launched from a height `h//5` into the pool. Determine her maximum airbone height `y` in terms of `h` and `theta`.

Explanation

Launch speed is sound from
`mg(4/5h)=1/2mv^2`
`v=sqrt(2g(4/5)h)impliesv_y=vsintheta`
The height y above the water (by conservation of energy for the child-earth system) is found from

`mgh=1/2m_y^2+mgh/5`
(since `1/2mv_x^2` is constant in projectile motion)
`y=(1)/(2g)v_y^2+h/5=(1)/(2g)v^2sin^2theta+h/5`
`y=(1)/(2g)[2g(4/5h)]sin^2theta+h/5=4/5hsin^2theta+h/5`