A stone is thrown vertically upward with an initial velocity of 40; m/s. Taking g=10m/s^2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Answers (1)

Initial velocity u = 40 m s-1

Acceleration a = -g = -10 m s-2

Final velocity at the highest point would be v = 0

Let the maximum height reached be s

As per the third equation of motion

\v^2-u^2=2as\ s=fracv^2-u^22a\ s=frac0^2-40^22	imes -10\ s=80m

The net displacement would be zero as the stone will return to the point from where it was thrown.

The total distance covered by the stone = 2s = 160 m

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