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Q. 15.     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?

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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=\frac{v^{2}-u^{2}}{2a}\\ s=\frac{0^{2}-40^{2}}{2\times -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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