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For an object thrown at $45 °$ to horizontal, the maximum height $(H)$ and horizontal range $(R)$ are related as :

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$R = 16 H$

$R = 8 H$

$R = 4 H$

$R = 2 H$

answer is C.

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Detailed Solution

$H = \frac{u^{2} \sin^{2} 45 °}{2 g} = \frac{u^{2}}{4 g} \Rightarrow R = \frac{u^{2}}{g} = 4 H$

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