    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.  Suggest Corrections  0      Similar questions  Related Videos   Jumping Off Cliffs
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