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 1.50 ft/s, how short will the packet fall?
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:
⇒
∴ 
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:
So, 
On solving this equation, we get:
T=2.99 sec
Distance at which packet will fall short = 288-35.81=192.19 ft.