An arrow is shot from a height of 1.5 m toward a cliff of height H . It is shot with a velocity of 30 m/s at an angle of 60º above the horizontal. It lands on the top edge of the cliff 4.0 s later.
(a) What is the height of the cliff?
(b) What is the maximum height reached by the arrow along its trajectory?
(c) What is the arrow’s impact speed just before hitting the cliff?
Solution:
Consider the following illustration:
Part A
We are required to solve for the value of H. We shall use the formula
\Delta \text{y}=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2or, we can also write the formula as
\text{y}-\text{y}_0=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2Substituting the given values, we have
\begin{align*}
\text{y}-\text{y}_0 &=\text{v}_{\text{0y}}\text{t}+\frac{1}{2}\text{at}^2 \\
\text{H}-1.5\:\text{m} & =\left(30\:\text{m/s}\:\sin 60^{\circ} \right)\left(4.0\:\text{s}\right)+\frac{1}{2}\left(-9.81\:\text{m/s}^2\right)\left(4.0\:\text{s}\right)^2 \\
\text{H}-1.5\:\text{m} & =25.44\:\text{m} \\
\text{H} & =25.44\:\text{m}+1.5\:\text{m} \\
\text{H} & =26.94\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}Part B
The maximum height of the projectile is given by the formula
\Delta \text{y}=\frac{\text{v}_{\text{0y}}^2}{2\text{g}}or the formula can be written as
\text{y}_{\text{max}}-\text{y}_{\text{0}}=\frac{\text{v}_{\text{0y}}^2}{-2\text{g}}Therefore, we have
\begin{align*}
\text{y}_{\text{max}}-\text{y}_{\text{0}} & =\frac{\text{v}_{\text{0y}}^2}{-2\text{g}} \\
\text{y}_{\text{max}}& =\frac{\text{v}_{\text{0y}}^2}{-2\text{g}}+\text{y}_{\text{0}} \\
\text{y}_{\text{max}}&=\frac{\left(\left(30\:\text{m/s}\right)\sin 60^{\circ} \right)^2}{-2\left(-9.81\:\text{m/s}^2\right)}+1.5\:\text{m} \\
\text{y}_{\text{max}}&=34.40\:\text{m}+1.5\:\text{m} \\
\text{y}_{\text{max}} & =35.90\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}
Part C
To solve for the final speed, we need the vertical and horizontal components when the arrow hits the cliff.
We are going to use the formula for acceleration along the vertical to solve for the final speed in the vertical direction. That is
\text{a}=\frac{\text{v}_{\text{y}}-\text{v}_{\text{0y}}}{\text{t}}Solving for vfy in terms of the other variables, we have
\begin{align*}
\text{v}_{\text{y}}&=\text{v}_{\text{0y}}+\text{at}\\
\text{v}_{\text{y}}&=\left(30\:\text{m/s}\right)\sin 60^{\circ} +\left(-9.81\:\text{m/s}^2\right)\left(4.0\:\text{s}\right) \\
\text{v}_{\text{y}}&=-13.25\:\text{m/s}
\end{align*}
Since we know that the horizontal component of the velocity does not change along the entire flight, we can equate the initial and final horizontal velocities. That is
\begin{align*}
\text{v}_{\text{x}}&=\text{v}_{\text{0x}}=\left(30\:\text{m/s}\right)\cos 60^{\circ} \\
\text{v}_{\text{x}}&=15\:\text{m/s}
\end{align*}
Therefore, the final speed is
\begin{align*}
\text{v}&=\sqrt{\text{v}_{\text{y}}^2+\text{v}_{\text{x}}^2} \\
\text{v}&=\sqrt{\left(-13.25\:\text{m/s}\right)^2+\left(15\:\text{m/s}\right)^2}\\
\text{v}&=20.01\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}