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Question

A stone is dropped from a cliff. Find its speed after the stone has fallen from 100 m. (Take g=9.8 ms−2).


A
9.8 ms1
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B
19.6 ms1
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C
44.2 ms1
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D
24 ms1
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Solution

The correct option is C 44.2 ms1

In case of free fall, the body would accelerate with 'g' (Therefore, a = g)
Given, initial velocity, u =0 ms1 (initially stone is at rest), acceleration due to gravity, g = 9.8 ms2 and height, h = 100 m.
Let the final velocity be 'v'.
In the case of free fall, the third equation of motion becomes , v2​ = u2​ + 2gh
v2​​ =​ 0 + 2 × 9.8 × 100
v2​​ = 1960
v = 1960 = 44.2 ms1


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