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Question

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.


A

36 feet

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B

12 feet

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C

0 feet the packet reaches his friend

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D

192 feet

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Solution

The correct option is D

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.


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