91Ó°ÊÓ

Substitute $x=156.25$in the given function,

$$\begin{gathered} h\left(156.25\right)=-\frac{32{\left(156.25\right)}^2}{{\left(100\right)}^2}+156.25 \\ =78.125 \end{gathered}$$

Therefore, the maximum height of the projectile is $78.125ft$.

When the projectile strikes the water, the height will be $0$. Then,

role="math" localid="1647363934493" $$\begin{gathered} 0=\frac{-32x^2}{{\left(100\right)}^2}+x \\ x\left(32x-10000\right)=0 \\ x=0,312.5 \end{gathered}$$

As the distance cannot be $0$, then the only possible value for x is $312.5$.

Therefore, the projectile will strike the ground at a horizontal distance of $312.5ft$from the firing point.

Plot the function for the interval $0≤x≤350$,

Consider the given question,

From the graph,

We can see how the maximum of the function is reached at the point$\left(156,78\right)$, while the projection will touch the ground for$x=313$.

We have to solve the quadratic equation $0.0032x^2-x+50=0$, use the quadratic formula,

$$\begin{gathered} x=\frac{1Â\pm \sqrt{1-0.64}}{0.0064} \\ x=250,62.5 \end{gathered}$$

Since the path of a projectile is a parabola, it is possible for the ball to reach a height of $50ft$twice. The second time the projectile reaches a height of $50ft$is after crossing the maximum height.

Therefore, the projectile has traveled $62.5ft,250ft$horizontally, when the height is$50ft$above the ground.