Answer
(a) The initial speed of the ball is
$$v_0=42.8 \text{ m/}\text{s}^2$$
(b) Height above the top of the fence is
$$H = 42 \text{ m}.$$
Work Step by Step
Let's wrtie down the equations for the horizontal and vertical displacement. We choose the coordinate system as shown in the figure. This will give us two equations in two unknowns ($v_0$ and $t$). And we will use the following initial conditions
y-component:
$a_y = -g = -9.81$ $m/s^2$,
$y_0=0.9$ $m$,
$v_{0y}=v_0\sin45^{\circ}$.
x-component:
$a_x=0,$
$x_0=0,$
$v_{0x}=v_0\sin45^{\circ}$.
So we have
$y=0.9+\dfrac{\sqrt{2}}{2}v_0t-\dfrac{9.8t^2}{2}$
$x=\dfrac{\sqrt{2}}{2}v_0t$
(a)
At time $t_{\text{end}}$ we know that $x(t_\text{end})=L=188$ $m$ and $y(t_\text{end})=0$.
$0=0.9+\dfrac{\sqrt{2}}{2}v_0t_{\text{end}}-\dfrac{9.8t_{\text{end}}^2}{2}$
$188=\dfrac{\sqrt{2}}{2}v_0t_{\text{end}}$
For $t_{\text{end}}$ we have
$t_{\text{end}}=\dfrac{2\cdot188}{\sqrt{2}v_0}$
and
$0 = 0.9 + \dfrac{\sqrt{2}}{2}v_0\cdot \dfrac{2\cdot188}{\sqrt{2}v_0}-\dfrac{9.8}{2}\left(\dfrac{2\cdot188}{\sqrt{2}v_0}\right)^2$
or
$v_0=\left(\dfrac{9.8\cdot188^2}{0.9+188}\right)^{1/2}$
$$v_0=42.8 \text{ m/s}$$
(b)
We will use the horizontal motion to find the time it takes the ball to reach the fence:
$116=\dfrac{\sqrt{2}}{2}\cdot 42.8 \cdot t$
or
$t = \dfrac{116 \text{m}}{30.3 \text{m/s}}=3.83 \text{ s}.$
Let's find the vertical displacement of the ball at this t
$y=0.9+\dfrac{\sqrt{2}}{2}\cdot42.8\cdot3.83-\dfrac{9.8}{2}\cdot3.83^2 \text{ m}$
$y= 45 \text{ m}$.
So Its height then above the top of the fence is
$H=45\text{ m} - 3\text{ m}=42 \text{ m}.$
