Hi Abby,

The important expression here is

\[s = u t + \frac12 a t^2 \]

where $s$ is the displacement, $u$ is the initial velocity, $t$ is time and $a$ is acceleration. You need to use this expression twice, once in the horizontal direction and again in the vertical direction.

In the horizontal the initial velocity is $u \cos\left(40^o\right)$ feet per second and since the acceleration in the horizontal direction is zero,

\[s = u \cos\left(40^o\right) t.\]

Let $T$ be the time in seconds when the ball hits the ground then at this time $s = 36$ feet hence

\[36 = u \cos\left(40^o\right) T \mbox{ or } T = \frac{36}{u \cos\left(40^o\right)}.\]

In the vertical direction, measuring upwards as positive, $a = -32$ feet per second squared and the initial velocity is $u \sin\left(40^o\right)$ and hence

\[s = u \sin\left(40^o\right) t - 16 t^{2}.\]

At the time $T$ seconds $s = -5 \frac{1}{12}$ feet and hence

\[5.08333 = u \sin\left(40^o\right) T - 16 T^{2}.\]

Substitute the expression for $T$ from the horizontal analysis and simplify. You should have a quadratic equation which you can solve for $u.$

I hope this helps,
Penny