Question

A person standing on the top of a cliff 171 ft high has to throw a packet to his friend standing on the ground 228 ft horizontally away. If he throws the packet directly aiming at the friend with a speed of 15.0 ft/s, how short will the packet fall?

Answer 1

Given:
Height (h) of the cliff = 171 ft
Horizontal distance from the bottom of the cliff = 228 ft
As per the question, the person throws the packet directly aiming to his friend at the initial speed (u) of 15.0 ft/s.


From the diagram, we can write:
$\tan \mathrm{\theta }=\frac{\mathrm{P}}{\mathrm{B}}=\frac{171}{228}$
$\Rightarrow \mathrm{\theta }={\tan }^{-1}\left(\frac{171}{228}\right)$
∴ θ = 37°
When the person throws the packet from the top of the cliff, it moves in projectile motion.
Let us take the reference axis at point A.
u is below the x-axis.
a = g = 32.2 ft/s² (Acceleration due to gravity)
Using the second equation of motion, we get:
$$\begin{gathered} y=u\sin \left(\theta \right)T+\frac{1}{2}g{T}^2 \\ y=171\mathrm{ft} \\ \theta =37° \\ g=32\mathrm{ft}/{\mathrm{s}}^2 \\ T=\mathrm{Time}\mathrm{of}\mathrm{flight} \\ 171=15\sin \left(37\right)T+\frac{1}{2}\times 32\times T^2 \\ \mathrm{On}\mathrm{solving}\mathrm{this}\mathrm{quadratic}\mathrm{equation}\mathrm{in}T,\mathrm{we}\mathrm{get}: \\ T=2.99\mathrm{s} \\ \mathrm{Range}=15\cos \left(37\right)\times 2.99=35.81\mathrm{ft} \\ \mathrm{Distance}\mathrm{by}\mathrm{which}\mathrm{the}\mathrm{packet}\mathrm{will}\mathrm{fall}\mathrm{short}=228-35.81=192.19\mathrm{ft} \end{gathered}$$