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Karnataka Board PUCPUC Science Class 11

A Ball is Dropped from a Height. If It Takes 0.200 S to Cross the Last 6.00 M before Hitting the Ground, Find the Height from Which It Was Dropped. Take G = 10 M/S2.

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Question

A ball is dropped from a height. If it takes 0.200 s to cross the last 6.00 m before hitting the ground, find the height from which it was dropped. Take g = 10 m/s2.

 
Short/Brief Note
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Solution

Given:
Distance travelled by the ball in 0.200 seconds = 6 m
Let:
Time, t = 0.200 s
Distance, s = 6 m
a = g = 10 m/s2 (Acceleration due to gravity)
Using the equation of motion, we get:

\[s = ut + \frac{1}{2}a t^2 \]

\[6 = u\left( 0 . 2 \right) + \frac{1}{2} \times 10 \times 0 . 04\]

\[ \Rightarrow u = \frac{5 . 8}{0 . 2} = 29 \text{ m } /s\]

Let h be the height from which the ball is dropped.
We have:
u = 0 and v = 29 m/s
Now,

\[h = \frac{v^2 - u^2}{2a}\] 

\[\Rightarrow h = \frac{{29}^2 - 0^2}{2 \times 10} = \frac{29 \times 29}{20} = 42 . 05 \text{ m } \]

∴ Total height = 42.05 + 6 = 48.05 m ≈ 48 m

 
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Chapter 3: Rest and Motion: Kinematics - Exercise [Page 52]

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HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 3 Rest and Motion: Kinematics
Exercise | Q 29 | Page 52

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