A baseball is hit at ground level. The ball reaches its maximum height above the ground level $3.0\ s$ after being hit. Then $2.5\ s$ after reaching its maximum height, the ball barely clears a fence that is $97.5\ m$ horizontally from where it was hit. Assume the ground is level.

1. What is the maximum height above the ground reached by the ball?

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We begin by setting up the kinematic equations for this problems. Here, we have:

x_0=0\ \ \ \ ;\ \ \ \ \ y_0=0\ \ \ \ \ ;\ \ \ \ \ v_{0x}=v_0{\mathrm{cos} \left({\theta }_0\right)\ }\ \ \ \ \ ;\ \ \ \ \ v_{0y}=v_0{\mathrm{sin} \left({\theta }_0\right)\ }

The kinematic equations are therefore:

\left\{ \begin{array}{c} v_x\left(t\right)=v_0{\mathrm{cos} \left({\theta }_0\right)\ } \\ v_y\left(t\right)=-gt+v_0{\mathrm{sin} \left({\theta }_0\right)\ } \end{array} \right.\ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c} x\left(t\right)=v_0{\mathrm{cos} \left({\theta }_0\right)\ }t \\ \displaystyle{y\left(t\right)=-\frac{1}{2}gt^2+v_0{\mathrm{sin} \left({\theta }_0\right)\ }t} \end{array} \right.

However in this problem we must proceed with caution as the initial velocity $v_0$ and the launch angle ${\theta }_0$ are unknown. This means that we cannot just start answering the classic questions — such as the maximum height of the projectile — without first determining $v_0$ and ${\theta }_0$ or the components $v_{0x}=v_0{\mathrm{cos} \left({\theta }_0\right)\ }$ and $v_{0y}=v_0{\mathrm{sin} \left({\theta }_0\right)\ }$ of the initial velocity (which equivalently take into account $v_0$ and ${\theta }_0$).

To derive the $y$-component $v_{0y}=v_0{\mathrm{si}\mathrm{n} \left({\theta }_0\right)\ }$, we write that the ball reaches its maximum height at a time $t_1=3.0\ s$ for which its vertical velocity cancels. We therefore have

\begin{aligned} v_y\left(t_1\right)=0\ \ \ \ &\Rightarrow \ \ \ \ -gt_1+v_0{\mathrm{sin} \left({\theta }_0\right)\ }=0 \\ &\Rightarrow \ \ \ \ \ v_0{\mathrm{sin} \left({\theta }_0\right)\ }=gt_1 \end{aligned}

This gives us the value of $v_0{\mathrm{sin} \left({\theta }_0\right)\ }$ since $t_1$ is known, and we then write that the maximum height is reached for $t_1$ and is equal to

\begin{aligned} y_{max}&=y\left(t_1\right) \\ &=-\frac{1}{2}gt^2_1+v_0{\mathrm{sin} \left({\theta }_0\right)\ }t_1 \\ &=-\frac{1}{2}gt^2_1+\left(gt_1\right)\cdot t_1 \\ &=-\frac{1}{2}\left(9.81\right)\cdot 3^2+3\cdot 9.8\cdot 3 \\ &\approx 44.05\ m \end{aligned}

where we used the value found for $v_0{\mathrm{sin} \left({\theta }_0\right)\ }$ from before. Thus, we conclude that the maximum height reached by the ball is equal to

\boxed{y_{max}=-\frac{1}{2}gt^2_1+\left(gt_1\right)\cdot t_1\approx 44.05\ m}

2. How high is the fence?

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Let us call $t_2$ the time at which the ball is at the top of the fence. At that point, we can write:

\left\{ \begin{array}{c} x\left(t_2\right)=D \ \ \ \ \ \ \\ y\left(t_2\right)=h_{fence} \end{array} \right.

We know that $t_2=3.0+2.5=5.5\ s$. The height of the fence being equal to the height of the ball at time $t_2$, we have:

h_{fence}=y\left(t_2\right)\ \ \ \ \Rightarrow \ \ \ \ \ h_{fence}=-\frac{1}{2}gt^2_2+v_0{\mathrm{sin} \left({\theta }_0\right)\ }t_2

Thus, recalling that $v_0{\mathrm{sin} \left({\theta }_0\right)\ }=gt_1$, the height of the fence is equal to

\boxed{h_{fence}=-\frac{1}{2}gt^2_2+\left(gt_1\right)\cdot t_2=-\frac{1}{2}\left(9.81\right){\left(5.5\right)}^2+3\cdot 9.81\cdot 5.5\approx 13.64\ m}

3. How far beyond the fence does the ball strike the ground?

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We label $d$ the distance between the fence and the point of impact of the ball. If $t_3$ is the time at which the ball hits the ground, then we may write:

\left\{ \begin{array}{c} x\left(t_3\right)=D+d \\ y\left(t_3\right)=0 \ \ \ \ \ \ \ \ \end{array} \right.

Using the second equation, we derive $t_3$ as follows:

\begin{aligned} y\left(t_3\right)=0\ \ \ \ &\Rightarrow \ \ \ \ -\frac{1}{2}gt^2_3+v_0{\mathrm{sin} \left({\theta }_0\right)\ }t_3=0 \\ &\Rightarrow \ \ \ \ \ t_3\left(-\frac{1}{2}gt_3+v_0{\mathrm{sin} \left({\theta }_0\right)\ }\right)=0 \\ &\Rightarrow \ \ \ \ \ t_3=0\ \ \ \ \ or\ \ \ t_3=\frac{2v_0{\mathrm{sin} \left({\theta }_0\right)\ }}{g} \end{aligned}

Thus, disregarding the solution $t_3=0\ s$, we have:

t_3=2\cdot 3\cdot \frac{9.81}{9.81}=6.0\ s

We may then use the first equation and write:

x\left(t_3\right)=v_0{\mathrm{cos} \left({\theta }_0\right)\ }t_3=D+d

This equation requires knowing what $v_0{\mathrm{cos} \left({\theta }_0\right)\ }$ is. We derive its value by considering that after $t_2=5.5\ s$, the ball has reached $D=97.5\ m$. Thus, we have

v_0{\mathrm{cos} \left({\theta }_0\right)\ }t_2=D\ \ \ \Rightarrow \ \ \ \ \ v_0{\mathrm{cos} \left({\theta }_0\right)\ }=\frac{D}{t_2}=\frac{97.5}{5.5}\approx 17.72\ m/s

The ball lands a distance $d$ behind the fence equal to

\boxed{d=v_0{\mathrm{cos} \left({\theta }_0\right)\ }t_3-D\approx 17.72\cdot 6-97.5\approx 8.82\ m}

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