When a ball is thrown up vertically with velocity V_0 , it reaches a maximum height of ' h '. If one wishes to triple the maximum height then the ball should be thrown with velocity
Text Solution
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At maximum height velocity is zero.
From equation of motion we have
v^{2} = u^{2} + 2gh
where v is final velocity, u is initial velocity. Since ball reaches maximum height, velocity at the highest point is zero. Therefore, we have
\(v = 0, u = v_0 \\ \Rightarrow v_0 = \sqrt{2gh}\)
when h' = 3h then
\(\mathbf{v_0'} = \sqrt{2g \times 3h} = \sqrt{3} \sqrt{2gh} = \sqrt{3} \mathbf{v_0}\)
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