A stone is thrown horizontally with a velocity ` 2 sqrt gh` from the top of a tower of height ` h`. It strikes the ground level through the foot of tower at a distance (x) from it . What is the value of 9x) ?

Explanation

Taking vertical downward motion for stone from starting point ot striking point on groud level, we have
` y_0 =0, y=h, u_y t -0, a_y =g, t=t 9say)`
` y=y_0 + u_y t + 1/23 a_y t^2`
:. ` h= 0 + 0 1/2 gt^2 or t= sqrt (2 h) /g`
Taking horixontal motion of stone from starting point to striking point on ground level. we have,
` x_0 = 0, x u_x =2 sqrt gh , a_x = 0, t =sqrt 2 h//g`
As ` x = x_0 + u_x t + 1/2 a_x t^2`
:. ` x =0 + 2 sqrt gh xx sqrt (2 h0/g + 1/2/ xx 0 xx t^2 =2 sqrt 2 h`.