A ball is thrown into the air at ground level. After $1.50\ s$, it’s displaced $35.0\ m$ horizontally and $45.0\ m$ vertically above its launch point. Find:

1. the initial speed

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First, we summarize the information provided in the following sketch

Second, we write the kinematic equations of the ball while it is in the air

\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.\ \ \ \ \ \ and\ \ \ \ \ \ \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.

Since $v_0$ and ${\theta }_0$ are both unknown and cannot easily be separated and solved for, recall that $v_{0x}=v_0{\mathrm{cos} \left({\theta }_0\right)\ }$ and $v_{0y}=v_0{\mathrm{sin} \left({\theta }_0\right)\ }$. In this case, solving for $v_{0x}$ and $v_{0y}$ is simpler as shown below.

At time $t_1=1.5\ s$ the ball is located at $x=x_1$ and we therefore solve

\begin{aligned} x\left(t_1\right)=x_1\ \ \ \ &\Rightarrow \ \ \ \ \ v_{0x}t_1=x_1 \\ &\Rightarrow \ \ \ \ \ v_{0x}=\frac{x_1}{t_1} \end{aligned}

The $x$-component of the initial velocity is equal to

\boxed{v_{0x}=\frac{x_1}{t_1}=\frac{35}{1.5}\approx 23.33\ m/s}

At time $t_1=1.5\ s$ the ball is located at $y=y_1$ and we therefore solve

\begin{aligned} y\left(t_1\right)=y_1\ \ \ \ \ &\Rightarrow \ \ \ \ \ -\frac{1}{2}gt^2_1+v_{0y}t_1=y_1 \\ &\Rightarrow \ \ \ \ \ \ v_{0y}=\frac{y_1+\frac{1}{2}gt^2_1}{t_1} \end{aligned}

The $y$-component of the initial velocity is equal to

\boxed{v_{0y}=\frac{y_1+\frac{1}{2}gt^2_1}{t_1}=\frac{45+\frac{1}{2}\left(9.8\right){\left(1.5\right)}^2}{1.5}=37.35\ m/s}

Thus, the initial speed of the ball is equal to

\boxed{v_0=\sqrt{v^2_{0x}+v^2_{0y}}\approx \sqrt{{23.33}^2+{37.35}^2}\approx 44.04\ m/s}

2. the initial angle above the horizontal

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The initial angle ${\theta }_0$ is given by

{\mathrm{tan} \left({\theta }_0\right)\ }=\frac{v_{0y}}{v_{0x}}\ \ \ \ \ \Rightarrow \ \ \ \ \ \ \boxed{{\theta }_0={{\mathrm{tan}}^{-1} \left(\frac{v_{0y}}{v_{0x}}\right)\ }\approx {{\mathrm{tan}}^{-1} \left(\frac{37.35}{23.33}\right)\ }\approx 58{}^\circ}

3. the maximum height reached

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The ball reaches its maximum height at $t=t_2$ such that its vertical height cancels (turn-around point) and therefore we write

\begin{aligned} v_y\left(t_2\right)=0\ \ \ \ &\Rightarrow \ \ \ \ -gt_2+v_0{\mathrm{sin} \left({\theta }_0\right)\ }=0 \\ &\Rightarrow \ \ \ \ \ t_2=\frac{v_0{\mathrm{sin} \left({\theta }_0\right)\ }}{g} \end{aligned}

The maximum height reached by the ball is therefore given by

\begin{aligned} y\left(t_2\right)&=-\frac{1}{2}gt^2_2+v_0{\mathrm{sin} \left({\theta }_0\right)\ }t_2 \\ \\ &=-\frac{1}{2}g\left(\frac{v^2_0{{\mathrm{sin}}^2 \left({\theta }_0\right)\ }}{g^2}\right)+\frac{v^2_0{{\mathrm{sin}}^2 \left({\theta }_0\right)\ }}{g} \\ \\ &=\frac{v^2_0{{\mathrm{sin}}^2 \left({\theta }_0\right)\ }}{2g} \\ \\ &\approx \frac{{44.04}^2\cdot {{\mathrm{sin}}^2 \left(58\right)\ }}{2\left(9.8\right)} \\ \\ &\approx 71.16\ m \end{aligned}

Thus, the maximum height reached by the ball is equal to

\boxed{y(t_2)=-\frac{1}{2}gt^2_2+v_0{\mathrm{sin} \left({\theta }_0\right)\ }t_2 \approx 71.16 \ m}

4. the total horizontal distance traveled when it finally returns to the ground

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Given the symmetry of the path, the ball returns to the ground at time $t_3=2t_2$. At that time, it has travelled a horizontal distance equal to

\begin{aligned} x\left(t_3\right)&=v_0{\mathrm{cos} \left({\theta }_0\right)\ }t_3 \\ \\ &=\frac{2v^2_0{\mathrm{sin} \left({\theta }_0\right)\ }{\mathrm{cos} \left({\theta }_0\right)\ }}{g} \\ \\ &=\frac{2\cdot {44.04}^2\cdot {\mathrm{sin} \left(58\right)\ }{\mathrm{cos} \left(58\right)\ }}{9.8} \\ \\ &\approx 177.88\ m \end{aligned}

Thus, the horizontal distance traveled by the ball when it finally returns to the ground is equal toThus, the horizontal distance traveled by the ball when it finally returns to the ground is equal to

\boxed{x\left(t_3\right)=v_0{\mathrm{cos} \left({\theta }_0\right)\ }t_3 \approx 177.88 \ m}

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