Question

A person standing on the top of a cliff, $171ft$ 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 $15.0ft/s$, how short will the packet fall? Give $g$ = $32ft/s^2$.

• A. 36 feet
• B. 12 feet
• C. 0 feet the packet reaches his friend
• D. 192 feet

Answer 1

The correct option is D

192 feet

In the given figure we can find $\theta$ by using trigonometry.

$tan\theta$ = $\frac{171}{228}$ = $\frac{3}{4}$

$\Rightarrow$ $\theta$ = ${37}^{\circ }$.

This means the ball was thrown at an angle ${37}^{\circ }$ to horizontal also its given that it was thrown with velocity 15 ft/s.

The dotted line shows how the trajectory of the ball thrown

So we know 2 D is nothing but 2 one dimensional motion.

Let's break the components:

$u_y$ = $-15sin{37}^{\circ }$ = $-15\times \frac{3}{5}$ = $-9ft/s$

$a_y$ = $-32ft/s^2$

$s_y$ = $-171ft$.

$\Rightarrow$ $s_y$ = $u_{y}t+\frac{1}{2}{a}_y{t}^2$

$-171$ = $-9t-\frac{32}{2}{t}^2$

$16t^2+9t-171=0$

$16t^2+57t-48t-171=0$

$\Rightarrow$ $t$ = $3sec$

We need to know in that time the projectile covered how much of horizontal distance.

So $u_x=15cos{37}^{\circ }$ = $15\times \frac{4}{5}$ = $12ft/s$

$a_x$ = $0$; $t$ = $3$

$s_x$ = $u_{x}t$ = $12\times 3$ = $36ft$

$\Rightarrow$ The packet went only 36 ft towards his friend

$\Rightarrow$ The packet fell short by $(228-36)$

= $192ft$.