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?

1
by avaniDey

2015-09-15T04:13:57+05:30

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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)

So the answer  is  d1 * d2 / [ d1 + d2 ]

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