A person standing on the top of a cliff 171 ft high needs to throw...

Let the point of projection be the origin $(0,0)$. The friend is on the ground at a horizontal distance of $228 \text{ ft}$ and a vertical depth of $171 \text{ ft}$, so their coordinates are $(228, -171)$.

Since the packet is aimed directly at the friend, the angle of projection $\theta$ below the horizontal is given by:

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

This gives $\sin \theta = \dfrac{3}{5}$ and $\cos \theta = \dfrac{4}{5}$.

The initial speed of the packet is $u = 15.0 \text{ ft/s}$. The horizontal and vertical components of the initial velocity are:

$u_x = u \cos \theta = 15 \times \dfrac{4}{5} = 12 \text{ ft/s}$

$u_y = -u \sin \theta = -15 \times \dfrac{3}{5} = -9 \text{ ft/s}$

Taking the acceleration due to gravity as $g = 32 \text{ ft/s}^2$, the vertical displacement $S_y$ is $-171 \text{ ft}$. Using the second equation of motion:

$S_y = u_y t - \dfrac{1}{2} g t^2$

$-171 = -9t - \dfrac{1}{2}(32)t^2$

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

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

Solving the quadratic equation for $t$:

$t = \dfrac{-9 \pm \sqrt{81 - 4(16)(-171)}}{2(16)}$

$t = \dfrac{-9 \pm \sqrt{81 + 10944}}{32} = \dfrac{-9 \pm \sqrt{11025}}{32}$

$t = \dfrac{-9 \pm 105}{32}$

Taking the positive root, the time of flight is:

$t = \dfrac{96}{32} = 3 \text{ s}$

The horizontal distance traveled by the packet in this time is:

$x = u_x t = 12 \times 3 = 36 \text{ ft}$

The distance by which the packet falls short of the friend is:

$\text{Distance short} = 228 - 36 = 192 \text{ ft}$