A ball falls under gravity from height $10 m$ with an initial velocity $u$ . It hits the ground, loses $50$ % of the energy in collision and it rises to the same height. What is the value of $u$ (in m/s) ?

7

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14

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Solution

The correct option is B $14$

Velocity at ground is V before collision.
$V^{2} - u^{2} = 2 g h$
$V^{2} = 2 g h + u^{2}$
after collision K.E is 50% velocity let $V^{1}$
$\frac{1}{2} m V^{2} = 2 \times (\frac{1}{2} m V^{1^{2}})$
$V^{1^{2}} = \frac{2 g h + u^{2}}{2}$
after height h velocity becomes zero.
$O - V^{1^{2}} = - 2 g h$
$\frac{2 g h + u^{2}}{2} = 2 g h$
$u^{2} = (4 g h - 2 g h) = 2 g h$
$u^{2} = 2 \times 10 \times 10 = 200$
$u \approx 14 m / s$

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A ball falls under gravity from height 10m with an initial velocity u. It hits the ground, loses 50% of the energy in collision and it rises to the same height. What is the value of u (in m/s) ?

The correct option is B 14Velocity at ground is V before collision.V2−u2=2ghV2=2gh+u2after collision K.E is 50% velocity let V112mV2=2×(12mV12)V12=2gh+u22after height h velocity becomes zero.O−V12=−2gh2gh+u22=2ghu2=(4gh−2gh)=2ghu2=2×10×10=200u≈14m/s

A ball falls under gravity from height $10 m$ with an initial velocity $u$ . It hits the ground, loses $50$ % of the energy in collision and it rises to the same height. What is the value of $u$ (in m/s) ?