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?

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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?

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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?

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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}

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