Q. 154.4( 138 Votes )

# A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking *g* = 10 m/s^{2}, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone when it falls back to the ground?

Answer :

Given,

Initial velocity, u = 40 m/s

Final velocity, v = 0 (As the stone stops)

Acceleration due to gravity, g = - 10 m/s^{2}

Height, h =?

v^{2} = u^{2} + 2gh

(0)^{2} = (40)2 + 2 (-10) × h

0 = 1600 – 20 h

20 h = 1600

h =

h = 80 m

The maximum height to which stone is thrown up is 80 m.

As, the stone is thrown up from the ground and after reaching to a maximum height of 80m it falls back to the ground. Therefore, the net displacement of the stone is zero.

The distance covered by the stone in reaching the maximum height is 80 m.

The stone will cover the same distance of 80 m in coming back to ground.

So, the total distance covered by the stone = 80 + 80

= 160 m

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