Question

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?

Answer 1

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 ]

Answer 2

Explanation:

Pls see the pic attached

Hope it helps!!