A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used? (b) What other angle gives the same range, and why would it not be used? (c) How long did this pass take?

Solution:

To illustrate the problem, consider the following figure:

Part A

We are given the 7-meter range, R, and the initial velocity, v_o, of the projectile. We have R=7.0 m, and v_o=12.0 m/s. To solve for the angle of the initial velocity, we will use the formula for range

\text{R}=\frac{\text{v}^{2}_{\text{o}}\sin 2\theta _{\text{o}}}{g}

Solving for θ_o in terms of the other variables, we have

\begin{align*}
\text{gR} & =\text{v}_{\text{o}}^2\sin 2\theta _{\text{o}} \\
\sin 2\theta _{\text{o}} & =\frac{\text{gR}}{\text{v}_{\text{o}}^2} \\
2\theta _\text{o} & =\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right) \\
\theta _\text{o} & =\frac{1}{2}\sin ^{-1}\left(\frac{\text{gR}}{\text{v}_{\text{o}}^2}\right) \\
\end{align*}

Substituting the given values, we have

\begin{align*}
\theta _\text{o} & =\frac{1}{2} \sin ^{-1}\left[\frac{\left(9.81\text{m/s}^2\right)\left(7.0\text{m}\right)}{\left(12.0\text{m/s}\right)^2}\right] \\

\theta _\text{o} & =14.2^{\circ}

\qquad \qquad{\color{DarkOrange} \left( \text{Answer} \right)} \\
\end{align*}

Part B

The other angle that would give the same range is actually the complement of the solved angle in Part A. The other angle,

\theta _o'=90^{\circ} -14.24^{\circ} =75.8^{\circ} \qquad \qquad{\color{DarkOrange} \left( \text{Answer} \right)} \\

This angle is not used as often, because the time of flight will be longer. In rugby that means the defense would have a greater time to get into position to knock down or intercept the pass that has the larger angle of release.

Part C

We can use the x-component of the motion to solve for the time of flight.

\Delta \text{x}=\text{v}_\text{x}\text{t}

We need the horizontal component of the velocity. We should be able to solve for the component since we are already given the initial velocity and the angle.

\text{v}_{\text{x}}=\left(12\:\text{m/s}\right)\cos 14.24^{\circ} =11.63\:\text{m/s}

Therefore, the total time of flight is

\begin{align*}
\text{t} & =\frac{\Delta \text{x}}{\text{v}_{\text{x}}} \\
\text{t} & =\frac{7.0\:\text{m}}{11.63\:\text{m/s}} \\
\text{t} & =0.60\:\text{s}

\qquad \qquad{\color{DarkOrange} \left( \text{Answer} \right)} \\
\end{align*}