Question

a projectile is launched from the top of a 20 - m tall cliff towards another cliff 140 - m away that is 50 - m tall, as shown below. if the projectile lands at the point labeled with the “x” 5 s after being launched:

• determine at what speed the projectile was launched. (10pts)
• determine the maximum height above the ground that the projectile will reach. (10pts)

Explanation:

Step1: Analyze vertical displacement

The vertical displacement $\Delta y = 50 - 20=30$ m. The initial vertical - velocity is $v_{0y}=v_0\sin40^{\circ}$ and the acceleration $a=-g=- 9.8$ m/s². The horizontal displacement $\Delta x = 140$ m and the initial horizontal - velocity is $v_{0x}=v_0\cos40^{\circ}$. In the horizontal direction, $\Delta x = v_{0x}t$, so $t=\frac{\Delta x}{v_{0x}}=\frac{140}{v_0\cos40^{\circ}}$. In the vertical direction, $\Delta y=v_{0y}t+\frac{1}{2}at^{2}$. Substitute $v_{0y}=v_0\sin40^{\circ}$, $a = - 9.8$ m/s² and $t=\frac{140}{v_0\cos40^{\circ}}$ into the vertical - displacement equation:
\[

$$\begin{align*} 30&=v_0\sin40^{\circ}\times\frac{140}{v_0\cos40^{\circ}}-\frac{1}{2}\times9.8\times(\frac{140}{v_0\cos40^{\circ}})^2\\ 30&=140\tan40^{\circ}-\frac{1}{2}\times9.8\times\frac{140^{2}}{v_0^{2}\cos^{2}40^{\circ}}\\ \frac{1}{2}\times9.8\times\frac{140^{2}}{v_0^{2}\cos^{2}40^{\circ}}&=140\tan40^{\circ}- 30 \end{align*}$$

\]

Step2: Solve for the initial velocity $v_0$

First, calculate $140\tan40^{\circ}-30\approx140\times0.839 - 30=117.46 - 30 = 87.46$. And $\cos40^{\circ}\approx0.766$. Then, $\frac{1}{2}\times9.8\times\frac{140^{2}}{v_0^{2}\times0.766^{2}}=87.46$.
\[

$$\begin{align*} \frac{4.9\times19600}{v_0^{2}\times0.587}&=87.46\\ v_0^{2}&=\frac{4.9\times19600}{87.46\times0.587}\\ v_0^{2}&=\frac{96040}{51.349}\\ v_0^{2}&\approx1870.3\\ v_0&\approx43.25\text{ m/s} \end{align*}$$

\]

Step3: Calculate the maximum height

The maximum - height formula is $H = h_0+\frac{v_{0y}^{2}}{2g}$, where $h_0 = 20$ m and $v_{0y}=v_0\sin40^{\circ}\approx43.25\times0.643\approx27.81$ m/s. Then $H = 20+\frac{27.81^{2}}{2\times9.8}=20+\frac{773.3}{19.6}\approx20 + 39.45=59.45$ m.

Answer:

• The speed at which the projectile was launched is approximately $43.25$ m/s.
• The maximum height above the ground that the projectile will reach is approximately $59.45$ m.