A football is thrown at an unknown speed and angle (although we know the vertical component of the initial velocity is upward). It then lands at a location $15\ m$ below the level at which it was thrown. Someone observes the football to be traveling at a speed of $25\ m/s$ in a direction $65{}^\circ $ below the horizontal just before striking the ground.

1. What was the initial speed of the football? At what angle was it thrown above the horizontal?

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We let $v_0$ denote the initial speed of the football and $v_{0x}$ and $v_{0y}$ denotes its $x$ and $y$ components respectively. As the ball travels through the air, its kinematic equations are given by

\left\{ \begin{array}{c} \displaystyle{x\left(t\right)=\frac{1}{2}a_xt^2+v_{0x}t+x_0=v_{0x}t} \\ \\ \displaystyle{y\left(t\right)=\frac{1}{2}a_yt^2+v_{0y}t+y_0=-\frac{1}{2}gt^2+v_{0y}t+h} \end{array} \right.\ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \ \left\{ \begin{array}{c} v_x\left(t\right)=v_{0x} \\ \\ v_y\left(t\right)=-gt+v_{0y} \end{array} \right.

where $h=15\ m$ is the initial height from which the football is thrown and $\displaystyle{\left{ \begin{array}{c}
a_x=0 \
a_y=-g \end{array}
\right.}$ because the football only experiences the acceleration due to gravity.

To solve for the components $v_{0x}$ and $v_{0y}$, we consider each direction separately.

Solving for $v_{0x}$:

The horizontal velocity $v_x\left(t\right)$ of the football remains constant throughout its flight because $a_x=0\ \ m/s^2$. Thus, we conclude that the $x$-component of the velocity upon impact is equal to the $x$-component of the initial velocity. In other words, we have

v_{0x}=v_f{\mathrm{cos} \left(\beta \right)\ }=25{\mathrm{cos} \left(65\right)\ }\approx 10.57\ \ m/s

Solving for $v_{0y}$:

Since we cannot determine the time of flight $t_f$ without knowing the initial speed and launch angle, we instead use the formula $v^2_{fy}=v^2_{0y}+2a_y\mathrm{\Delta }y$ between the launch point and the point where the football hits the ground. We then derive $v_{0y}$ as follows

\begin{aligned} v^2_{fy}=v^2_{0y}+2a_y\mathrm{\Delta }y\ \ \ \ \ &\Rightarrow \ \ \ \ \ v^2_f{{\mathrm{sin}}^2 \left(\beta \right)\ }=v^2_{0y}+2\cdot \left(-g\right)\cdot \left(-h\right) \\ \\ &\Rightarrow \ \ \ \ \ v^2_{0y}=v^2_f{{\mathrm{sin}}^2 \left(\beta \right)\ }-2gh \\ \\ &\Rightarrow \ \ \ \ \ v_{0y}=\sqrt{v^2_f{{\mathrm{sin}}^2 \left(\beta \right)\ }-2gh} \end{aligned}

The component $v_{0y}$ of the initial velocity is therefore equal to

v_{0y}=\sqrt{v^2_f{{\mathrm{sin}}^2 \left(\beta \right)\ }-2gh}=\sqrt{{25}^2{{\mathrm{sin}}^2 \left(65\right)\ }-2\cdot 9.8\cdot 15}\approx 14.81\ m/s

The initial speed $v_0$ with which the football was launched is therefore equal to

\boxed{v_0=\sqrt{v^2_{0x}+v^2_{0y}}=\sqrt{{10.57}^2+{14.81}^2}\approx 18.20\ m/s}

and the launch angle $\theta $ is equal to

{\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{14.81}{10.57}\right)\ }\approx 54.48{}^\circ}

2. How much time did the ball spend in the air?

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To find the time of flight $t_f$, we solve $y\left(t_f\right)=0$ as follows

\begin{aligned} y\left(t_f\right)=0\ \ \ \ &\Rightarrow \ \ \ \ \ -\frac{1}{2}gt^2_f+v_{0y}t_f+h=0 \\ \\ &\Rightarrow \ \ \ \ \ \ gt^2_f-2v_{0y}t_f-2h=0 \\ \\ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c} \displaystyle{t_f=\frac{2v_{0y}-\sqrt{4v^2_{0y}+8gh}}{2g}\ < 0} \\ \\ \displaystyle{t_f=\frac{2v_{0y}+\sqrt{4v^2_{0y}+8gh}}{2g}\ > 0} \end{array} \right. \end{aligned}

Neglecting the first root of the quadratic which is negative, we conclude that the time the football spent in the air is equal to

\boxed{t_f=\frac{2v_{0y}+\sqrt{4v^2_{0y}+8gh}}{2g}=\frac{2\cdot 14.81+\sqrt{4\cdot {14.81}^2+8\cdot 9.8\cdot 15}}{2\cdot 9.8}\approx 3.82\ s}

Note: we could also have solved using $v_y\left(t_f\right)=-v_f{\mathrm{sin} \left(\beta \right)\ }$ which yields

\begin{aligned} v_y\left(t_f\right)=-v_f{\mathrm{sin} \left(\beta \right)\ }\ \ \ \ &\Rightarrow \ \ \ -gt_f+v_0{\mathrm{sin} \left(\theta \right)\ }=-v_f{\mathrm{sin} \left(\beta \right)\ } \\ \\ &\Rightarrow \ \ \ \ \ t_f=\frac{v_0{\mathrm{sin} \left(\theta \right)\ }+v_f{\mathrm{sin} \left(\beta \right)\ }}{g} \\ \\ &\Rightarrow \ \ \ \ \ t_f=\frac{18.20{\mathrm{sin} \left(54.48\right)\ }+25{\mathrm{sin} \left(65\right)\ }}{9.8}\approx 3.82\ s \end{aligned}

3. How far horizontally did the football travel?

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In the time $t_f$, the ball travels a horizontal distance $D$ equal to

\boxed{D=x\left(t_f\right)=v_{0x}t_f=10.57\cdot 3.82\approx 40.37\ \ m}

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