Answer
Please see the work below.
Work Step by Step
(a) We can find the required initial speed as follows:
$v_y^2=u_y^2-2gh$
We plug in the known values to obtain:
$0=u_y^2-2(9.8m/s^2)(61.5m)$
This simplifies to:
$u_y=34.71m/s$
(b) We know that the horizontal range from the top is given as
$x=v\sqrt{\frac{2h}{g}}$
We plug in the known values to obtain:
$x=24.71\sqrt{\frac{2\times 61.5}{9.8}}$
$x=122.96m$
and the maximum height from the ground is given as
$h=\frac{v_{\circ}^2}{g}$
We plug in the known values to obtain:
$h=\frac{(34.71)^2}{9.8}$
$h=122.96m$
Thus, the maximum height is the same from the top and the ground.
