The maximum distance to which a man can throw a ball by projecting it horizontally from a height h is h. The maximum distance to which he can throw it vertically up is

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h/4

h/2

answer is D.

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Detailed Solution

R = h in horizontal projection from certain height 'h'
Then $u \sqrt{\frac{2 h}{g}} = h \Rightarrow u = \sqrt{\frac{gh}{2}}$
$h_{vetical} = \frac{u^{2}}{2 g} = \frac{gh / 2}{2 g} = \frac{h}{4}$

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