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?

#### Solution

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:

\[\tan \theta = \frac{P}{B} = \frac{171}{228}\]

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/s

^{2}(Acceleration due to gravity)

Using the second equation of motion, we get:

\[y = u\sin\left( \theta \right)T + \frac{1}{2}g T^2 \]

\[y = 171 ft\]

\[\theta = 37^\circ\]

\[g = 32 ft/ s^2 \]

\[T = \text{ Time of flight } \]

\[171 = 15\sin\left( 37 \right)T + \frac{1}{2} \times 32 \times T^2 \]

\[\text{ On solving this quadratic equation in T, we get } : \]

\[T = 2 . 99 s\]

\[\text{ Range } = 15\cos\left( 37 \right) \times 2 . 99 = 35 . 81 ft\]

\[\text{ Distance by which the packet will fall short } = 228 - 35 . 81 = 192 . 19 ft\]