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A particle projected from ground at angle 45 degrees with horizontal from distance 'd1' from the foot of a pole and just after touching the top of pole it

falls on ground at a distance 'd2' from pole on other side, then what is the height of pole?

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Two dimensional projectile, :

x = u cosФ * t = u t/√2 as Ф = π/4 and, y = u sinФ t - 1/2 g t² = u/√2 - 1/2 g t²

y = x - g x² / u² as tanФ = 1 and cosФ =1/√2

let us take the coordinates of the point of projection as, (0, 0). when x = d1, y = h = height o f the pole. so h = d1 - g d1² / u² -- (1)

when x = d1 + d2, y = 0, hence, so 0 = (d1+d2) - g (d1 + d2)² / u² => u² = g (d1 + d2) --- (2)

substituting this (2) in (1) we get: h = d1 - d1² /(d1+d2) = d1 * d2 / (d1 + d2)