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P02G2016 – Ball Thrown in the Air

Ball Thrown in the Air

A ball is thrown in the air. When it is at a height of $35.0\ m$ above the ground, the ball is traveling at a speed of $25.0\ m/s$ at an angle of $15{}^\circ $ above the horizontal.

1. How fast was it initially thrown, and at what angle above the horizontal?

View answer

We first summarize the information of the problem in the figure below

The ball is hit from an initial position $\displaystyle{\left\{ \begin{array}{c}
x_0=0 \\
y_0=0 \end{array}
\right.}$ 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.

Since $v_0$ and $\theta $ are both unknown and cannot easily be separated and solved for, recall that $v_{0x}=v_0{\mathrm{cos} \left(\theta \right)\ }$ and $v_{0y}=v_0{\mathrm{sin} \left(\theta \right)\ }$. In this case, solving for $v_{0x}$ and $v_{0y}$ is simpler than directly solving for $v_0$ and $\theta $. This technique is relatively common in problems where both the initial speed and the launch angle are unknown.

Solving for $v_{0x}$:

The horizontal component of the ball’s velocity is constant throughout its motion because there is no acceleration in the $x$-direction. Thus, we conclude that the $x$-component of the velocity at height $H=35\ m$ — which is given by $v_1{\mathrm{cos} \left({\theta }_1\right)\ }$ — is equal to the $x$-component of the initial velocity $v_{0x}$ and we therefore write

\boxed{v_{0x}=v_1{\mathrm{cos} \left({\theta }_1\right)\ }=25\cdot {\mathrm{cos} \left(15\right)\ }\approx 24.15\ m/s}

Solving for $v_{0y}$:

The vertical component of the ball’s velocity at point 1, $v_{1y}=v_1{\mathrm{sin} \left({\theta }_1\right)\ }$, is related to the vertical component of its initial velocity $v_{0y}$ and the height $H$ at point 1 through the formula

v^2_{1y}=v^2_{0y}+2a_y\mathrm{\Delta }y

Solving for $v_{0y}$ then yields

\begin{aligned}
v^2_{1y}=v^2_{0y}+2a_y\mathrm{\Delta }y\ \ \ \ &\Rightarrow \ \ \ \ \ v^2_{1y}=v^2_{0y}+2\left(-g\right)\left(H-0\right) \\
\\
&\Rightarrow \ \ \ \ \ v_{0y}=\sqrt{v^2_1{{\mathrm{sin}}^2 \left({\theta }_1\right)\ }+2gH}
\end{aligned}

Thus, the vertical component of the ball’s initial velocity is equal to

\boxed{v_{0y}=\sqrt{v^2_1{{\mathrm{sin}}^2 \left({\theta }_1\right)\ }+2gH}=\sqrt{{25}^2{{\mathrm{sin}}^2 \left(15\right)\ }+2\left(9.8\right)\left(35\right)}\approx 26.98\ m/s}

The initial speed $v_0$ of the ball can then derived from its components and is equal to

\boxed{v_0=\sqrt{v^2_{0x}+v^2_{0y}}=\sqrt{{24.15}^2+{26.98}^2}\approx 36.21\ m/s}

Similarly, we derive the launch angle $\theta $ as follows

{\mathrm{tan} \left(\theta \right)\ }=\frac{v_{0y}}{v_{0x}}\ \ \ \ \Rightarrow \ \ \ \ \ \boxed{{\theta }={{\mathrm{tan}}^{-1} \left(\frac{v_{0y}}{v_{0x}}\right)\ }={{\mathrm{tan}}^{-1} \left(\frac{26.98}{24.15}\right)\ }\approx 48.17{}^\circ}

2. How long does it take to reach a height of $35.0\ m$ above the ground after being launched?

View answer

The ball reaches a height of $35.0\ m$ when its vertical velocity equals $v_1{\mathrm{sin} \left({\theta }_1\right)\ }$. Therefore, we solve:

\begin{aligned}
v_y\left(t_1\right)=v_1{\mathrm{sin} \left({\theta }_1\right)\ }\ \ \ \ &\Rightarrow \ \ \ \ \ -gt_1+v_0{\mathrm{sin} \left(\theta \right)\ }=v_1{\mathrm{sin} \left({\theta }_1\right)\ } \\
\\
&\Rightarrow \ \ \ \ \ t_1=\frac{v_0{\mathrm{sin} \left(\theta \right)\ }-v_1{\mathrm{sin} \left({\theta }_1\right)\ }}{g}
\end{aligned}

The time it takes the ball to reach a height of $35.0\ m$ above the ground is equal to

\boxed{t_1=\frac{v_0{\mathrm{sin} \left(\theta \right)\ }-v_1{\mathrm{sin} \left({\theta }_1\right)\ }}{g}\approx \frac{36.21\cdot {\mathrm{sin} \left(48.17\right)\ }-25\cdot {\mathrm{sin} \left(15\right)\ }}{9.8}\approx 2.1\ s}

3. At that point, how far is it horizontally from its initial launch point?

View answer

After a time $t_1\approx 2.1\ s$, the ball has covered a horizontal distance equal to

\boxed{x\left(t\right)=v_0{\mathrm{cos} \left(\theta \right)\ }t_1\approx 36.21\cdot {\mathrm{cos} \left(48.17\right)\ }\cdot 2.1\approx 50.71\ m}