When a ball is thrown up vertically with velocity $v_0$, it reaches a maximum height of '$h$'. If one wishes t

Question

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:

Accepted Answer

1. **Identify the Relationship:** For a body thrown vertically upwards with initial velocity $u$, the maximum height $H$ reached is given by the kinematic equation $v^2 = u^2 + 2as$. At maximum height, final velocity $v=0$ and acceleration $a = -g$.
   $$ 0 = u^2 - 2gH \implies u = \sqrt{2gH} $$
   This shows that the initial velocity is directly proportional to the square root of the maximum height ($u \propto \sqrt{H}$) .
2. **Apply the Condition:**
   - Case 1: Initial velocity $= v_0$, Max height $= h$. So, $v_0 \propto \sqrt{h}$.
   - Case 2: New max height $h' = 3h$. Let the new velocity be $v'$.
   $$ v' \propto \sqrt{3h} = \sqrt{3} 	imes \sqrt{h} $$
3. **Determine New Velocity:** Substituting $v_0$ for $\sqrt{h}$ (proportionality constant cancels out):
   $$ v' = \sqrt{3} v_0 $$

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