Answer
The ball reaches a maximum height of $~~76~m~~$ above the ground.
Work Step by Step
The initial height of the ball is $30~m$ above the ground.
The initial velocity is $(10~m/s+20~m/s)$ which is $30~m/s$
We can find the time to reach maximum height:
$v_f = v_0+at$
$t = \frac{v_f-v_0}{a}$
$t = \frac{0-30~m/s}{-9.8~m/s^2}$
$t = 3.06~s$
We can find the maximum height:
$y = y_0+v_0~t+\frac{1}{2}at^2$
$y = (30~m)+(30~m/s)(3.06~s)+\frac{1}{2}(-9.8~m/s^2)(3.06~s)^2$
$y = 76~m$
The ball reaches a maximum height of $~~76~m~~$ above the ground.
