A ball is projected vertically upwards. Its speed at half of maximum height is $20 m / s$ . The maximum height attained by it is [Take $g = 10 m / s^{2}$ ]

35 m

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15 m

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25 m

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40 m

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Solution

The correct option is D $40 m$
$(D)$
using : $v^{2} - u^{2} = 2 a s$
at max height: $v = 0, \Rightarrow u = \sqrt{2 g h_{m a x}}$
given: $s = \frac{h_{m a x}}{2}, v = 20$
using : $v^{2} - u^{2} = 2 a s$
$20^{2} - (\sqrt{2 g h_{m a x}})^{2} = - 2 g (\frac{h_{m a x}}{2})$
$400 = g h_{m a x}$
$h_{m a x} = 40$

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