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 at the friend wuth a speed of 15.0 ft/s, how short will the packet fall ?

Answer 1

Given, tan $\theta =\frac{171}{228}$

$\Rightarrow \theta =ta{n}^{-1}\left(\frac{171}{228}\right)$

The motion of projectile (i,e. the packed) is from A. Take reference axis at a.

$\therefore \theta ={37}^{0}a{s}^{\prime }{u}^{\prime }isbelowx=axis$

u = 15 ft/s, g = 32.2 $ft/s^2$, y = -171 ft

From y = y = $tan\theta x-\frac{x^{2}gse{c}^2\theta }{2u^2}$

$-171=-\left(0.7536\right)x-frac{x}^{2}g\left(1.568\right)2\left(225\right)$

= $0.1125x^2-0.7536x-171=0$

$\Rightarrow$ x = 35.78 ft (can be calculated)

Horizontal range covered by the packet is 35.78 ft.

So, the packet will fall

= 228 - 35.78

= 192 ft short of his friend.