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A person throws the ball with velocity v from top of building in vertically upward direction the ball reaches the ground with the speed of 3v then the height of buliding is ?

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Solution

Let, ‘h’ be the height of the building which is also the net displacement of the ball. Initial velocity is ‘v’ (upwards), final velocity is ‘2v’ (downwards), and acceleration due to gravity is ‘g = 9.8 m/s²’ (downwards).
Using,
v² = u² + 2as
=> (3v)² = v² + 2gh
=> h = 4v²/g

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