A stone is dropped from a cliff. Find its speed after the stone has fallen from 100 m. (Take g=9.8 ms−2).
In case of free fall, the body would accelerate with 'g' (Therefore, a = g)
Given, initial velocity, u =0 ms−1 (initially stone is at rest), acceleration due to gravity, g = 9.8 ms−2 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 ms−1