If a baseball player can throw a ball at maximum distance = d over a ground, the maximum vertical height to which he can throw it, will be (Ball have same initial speed in each case) -

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TestHub · Accepted Answer

To find the maximum horizontal distance $d$ , the projection angle is $4 5^{\circ}$ .

$d = \frac{u ^{2} sin ( 2 \times 4 5 ^{\circ} )}{g} = \frac{u ^{2}}{g}$

For maximum vertical height $H$ , the projection angle is $9 0^{\circ}$ .

$H = \frac{u ^{2} sin ^{2} ( 9 0 ^{\circ} )}{2 g} = \frac{u ^{2}}{2 g}$

Comparing $d$ and $H$ : $H = \frac{1}{2} (\frac{u ^{2}}{g}) = \frac{d}{2}$ .

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