



 
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,




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