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/s2 (Acceleration due to gravity)
Using the second equation of motion, we get:
\[y = u\sin\left( \theta \right)T + \frac{1}{2}g T^2 \]
\[y = 171 ft\]
\[\theta = 37^\circ\]
\[g = 32 ft/ s^2 \]
\[T = \text{ Time of flight } \]
\[171 = 15\sin\left( 37 \right)T + \frac{1}{2} \times 32 \times T^2 \]
\[\text{ On solving this quadratic equation in T, we get } : \]
\[T = 2 . 99 s\]
\[\text{ Range } = 15\cos\left( 37 \right) \times 2 . 99 = 35 . 81 ft\]
\[\text{ Distance by which the packet will fall short } = 228 - 35 . 81 = 192 . 19 ft\]
