The batter hits the baseball with an initial velocity ${\overrightarrow{v}}_0$ and at an angle $\theta =45{}^\circ $. The ball’s initial height is $h_0=1.2\ m$ and it reaches a maximum height $y_{max}=24.15\ m$. An outfielder is initially standing at a distance $x_A=81\ m$ from the origin. The outfielder can catch a ball up to a height of $3\ m$ if he jumps vertically. He can catch a ball up to a height of $2m$ if he is standing or running at constant speed $v_1=5\ m/s$.

1. Setup the kinematic equations of the ball’s motion. What is the unknown? Solve for it.

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The ball is hit from an initial position $\displaystyle{\left\{ \begin{array}{c}
x_0=0 \\
y_0=h_0 \end{array}
\right.}$ and the components of the initial velocity ${\overrightarrow{v}}_0$ and acceleration (due to gravity) are

\left\{ \begin{array}{c} v_{0x}=v_0{\mathrm{cos} \left(\theta \right)\ } \\ v_{0y}=v_0{\mathrm{sin} \left(\theta \right)\ } \end{array} \right.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c} a_x=0 \\ a_y=-g \end{array} \right.

The kinematic equations describing the flight of the ball are therefore given by

\ \left\{ \begin{array}{c} x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }t \\ \\ \displaystyle{y\left(t\right)=-\frac{1}{2}gt^2+v_0{\mathrm{sin} \left(\theta \right)\ }t+h_0} \end{array} \right.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c} v_x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ } \\ \\ v_y\left(t\right)=-gt+v_0{\mathrm{sin} \left(\theta \right)\ } \end{array} \right.

In this case, the initial speed $v_0$ is unknown. We can solve for it however by considering the initial position where the ball is hit and the maximum height that it reaches as shown below

Recalling that $v_y=0$ at the maximum height, we can relate the $y$-component of the initial velocity to the maximum height $y_{max}\ $by writing the following and then solving for $v_0$.

\begin{aligned} v^2_{fy}=v^2_{0y}+2a_y\mathrm{\Delta }y\ \ \ \ \ &\Rightarrow \ \ \ \ \ 0^2=v^2_0{{\mathrm{sin}}^2 \left(\theta \right)\ }+2\cdot \left(-g\right)\cdot \left(y_{max}-h_0\right) \\ \\ &\Rightarrow \ \ \ \ \ v^2_0{{\mathrm{sin}}^2 \left(\theta \right)\ }=2g\left(y_{max}-h_0\right) \\ \\ &\Rightarrow \ \ \ \ \ v_0=\sqrt{\frac{2g\left(y_{max}-h_0\right)}{{{\mathrm{sin}}^2 \left(\theta \right)\ }}} \end{aligned}

Thus, we conclude that the baseball is initially launched with a speed $v_0$ equal to

\boxed{v_0=\sqrt{\frac{2g\left(y_{max}-h_0\right)}{{{\mathrm{sin}}^2 \left(\theta \right)\ }}}=\sqrt{\frac{2\cdot 9.8\cdot \left(24.15-1.2\right)}{{{\mathrm{sin}}^2 \left(45\right)\ }}}\approx 30\ m/s}

2. Where along $x$ does the ball reach its maximum height $y_{max}$?

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The ball reaches its maximum heights for a time $t_1$ such that $v_y\left(t_1\right)=0$ and we first solve for $t_1$ as follows

\begin{aligned} v_y\left(t_1\right)=0\ \ \ \ \ &\Rightarrow \ \ \ \ \ -gt_1+v_0{\mathrm{sin} \left(\theta \right)\ }=0 \\ \\ &\Rightarrow \ \ \ \ \ t_1=\frac{v_0{\mathrm{sin} \left(\theta \right)\ }}{g}\approx 2.16\ s \end{aligned}

We then derive the distance $x_{max}$ by computing $x\left(t_1\right)$ which yields

\boxed{x_{max}=x\left(t_1\right)=v_0{\mathrm{cos} \left(\theta \right)\ }t_1=\frac{v^2_0{\mathrm{sin} \left(\theta \right)\ }{\mathrm{cos} \left(\theta \right)\ }}{g}=\frac{{30}^2\cdot {\mathrm{sin} \left(45\right)\ }{\mathrm{cos} \left(45\right)\ }}{9.8}\approx 45.92\ m}

3. If the outfielder jumps vertically, can he catch the ball with it is at point $A$ (assume it is directly above him)?

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The outfielder will catch the ball at $A$ if the height of the ball at that point is less than $3\ m$ above the ground. To derive the height of the ball at point $A$, we first derive the time $t_A$ it takes the ball to reach point $A$ by solving by solving $x\left(t_A\right)=x_A$ which yields

\begin{aligned} x\left(t_A\right)=x_A\ \ \ \ &\Rightarrow \ \ \ \ \ v_0{\mathrm{cos} \left(\theta \right)\ }t_A=x_A \\ \\ &\Rightarrow \ \ \ \ \ t_A=\frac{x_A}{v_0{\mathrm{cos} \left(\theta \right)\ }} \\ \\ &\Rightarrow \ \ \ \ \ t_A=\frac{x_A}{v_0{\mathrm{cos} \left(\theta \right)\ }} \end{aligned}

The height of the ball at that moment is equal then given by

\begin{aligned} y_A&=y\left(t_A\right) \\ \\ &=-\frac{1}{2}gt^2_A+v_0{\mathrm{sin} \left(\theta \right)\ }t_A+h_0 \\ \\ &=-\frac{1}{2}g{\left(\frac{x_A}{v_0{\mathrm{cos} \left(\theta \right)\ }}\right)}^2+v_0{\mathrm{sin} \left(\theta \right)\ }\cdot \frac{x_A}{v_0{\mathrm{cos} \left(\theta \right)\ }}+h_0 \\ \\ &=-\frac{gx^2_A}{2v^2_0{{\mathrm{cos}}^2 \left(\theta \right)\ }}+{\mathrm{tan} \left(\theta \right)\ }x_A+h_0 \\ \\ &=-\frac{9.8\cdot {81}^2}{2\cdot {30}^2\cdot {{\mathrm{cos}}^2 \left(45\right)\ }}+{\mathrm{tan} \left(45\right)\ }\cdot 81+1.2 \\ \\ &\approx \boxed{10.76\ m} \end{aligned}

Therefore, we conclude that the outfielder will not be able to catch the ball at point $x_A$ even if he jumps since the height of the ball exceeds $3\ m$.

4. If the outfielder runs to catch the ball at point $B$, how long can he wait after the ball is hit before he must start running?

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We will first determine the time it takes the ball to reach $B$ and then determine how long it takes the outfielder to get from $A$ to $B$.

The ball reaches $B$ when its position equals $x_B$ which occurs at time $t$ such that $x\left(t\right)=x_B$. Thus, we solve for $t$ as follows

\begin{aligned} x\left(t\right)=x_B\ \ \ \ &\Rightarrow \ \ \ \ \ v_0{\mathrm{cos} \left(\theta \right)\ }t=x_B \\ \\ &\Rightarrow \ \ \ \ \ t=\frac{x_B}{v_0{\mathrm{cos} \left(\theta \right)\ }} \\ \\ &\Rightarrow \ \ \ \ \ \boxed{t=\frac{91}{30\cdot {\mathrm{cos} \left(45\right)\ }}\approx 4.29\ s} \end{aligned}

and conclude that it takes the ball $4.29\ s$ to reach point $B$.

We then consider the time it takes the outfielder to reach point $B$ which is located $10\ m$ away from point $A$. Since he can run at constant speed $v_1=5\ m/s$, we conclude that it takes him $2\ s$ to reach point $B$ and that he must therefore start running $2.29\ s$ after the ball is hit to reach point $B$ and catch the ball.

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