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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. - Physics

Short Note

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?

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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:

\[\tan \theta = \frac{P}{B} = \frac{171}{228}\]

\[\Rightarrow \theta = \tan^{- 1} \left( \frac{171}{228} \right)\]
∴ θ = 37°
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:

\[y = u\sin\left( \theta \right)T + \frac{1}{2}g T^2 \]

\[y = 171 ft\]

\[\theta = 37^\circ\]

\[g = 32 ft/ s^2 \]

\[T = \text{ Time of flight } \]

\[171 = 15\sin\left( 37 \right)T + \frac{1}{2} \times 32 \times T^2 \]

\[\text{ On solving this quadratic equation in T, we get } : \]

\[T = 2 . 99 s\]

\[\text{ Range }  = 15\cos\left( 37 \right) \times 2 . 99 = 35 . 81 ft\]

\[\text{ Distance by which the packet will fall short }  = 228 - 35 . 81 = 192 . 19 ft\] 

  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 3 Rest and Motion: Kinematics
Q 37 | Page 52
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