CBSE Class 9 Science Question 14 of 33

Gravitation — Question 14

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Question

Question 14

A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

Answer

Given,

Initial velocity (u) = 0

Tower height = total distance = 19.6 m

g = 9.8 ms-2

Final velocity (v) = ?

According to the third equation of motion,

v2 = 2gs + u2

Substituting we get,

v2 = (2 x 9.8 x 19.6) + 0

v = 384.16\sqrt{384.16}

v = 19.6 ms-1

Hence, final velocity = 19.6 ms-1