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A body projected obliquely at 45° to horizontal just crosses a wall at a distance ‘a’ from point of projection and falls at a distance ‘b’ on the other side of wall. The maximum height above the wall that it can reach is

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$\frac{ab}{a + b}$

$\frac{a + b}{2}$

$\frac{ab}{a - b}$

$\frac{{(a - b)}^{2}}{4 (a + b)}$

answer is D.

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Detailed Solution

$h = x tanθ (1 - \frac{x}{R}) & R tanθ = 4 H$

$∆ h = H - h = \frac{a + b}{4} - a (1 - \frac{a}{a + b})$

$∆ h = H - h = \frac{{(a - b)}^{2}}{4 (a + b)}$

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