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? Give g = 32 ft/s2.
192 feet
In the given figure we can find θ by using trigonometry.
tan θ = 171228 = 34
⇒ θ = 37∘.
This means the ball was thrown at an angle 37∘ to horizontal also its given that it was thrown with velocity 15 ft/s.
The dotted line shows how the trajectory of the ball thrown
So we know 2 D is nothing but 2 one dimensional motion.
Let's break the components:
uy = −15 sin 37∘ = −15 × 35 = −9 ft/s
ay = −32 ft/s2
sy = −171 ft.
⇒ sy = uyt + 12ayt2
−171 = −9t − 322t2
16t2 + 9t − 171 = 0
16t2 + 57t − 48t − 171 = 0
⇒ t = 3 sec
We need to know in that time the projectile covered how much of horizontal distance.
So ux = 15 cos37∘ = 15 × 45 = 12 ft/s
ax = 0; t = 3
sx = uxt = 12 × 3 = 36 ft
⇒ The packet went only 36 ft towards his friend
⇒ The packet fell short by (228 − 36)
= 192ft.