A body is thrown horizontally from the top of a tower and strikes the ground after three seconds at an angle of `45^@` with the horizontal. Find the height of the tower and the speed with which the body was projected. (Take ` g = 9.8 m//s^2`)

Explanation

Let (u) be the horizontal velocity of projected body from the top of a tower. The intial vetical velocity of th body is zero i.e. ` u_y =0 `. Let the body reach the ground after ` 3` seconds with horizontal component velocity ` v-x ( = u) and vertical component velocity ` v_y`.
Taking vetical downward motion ofthe body from top of tower to ground. we have,
`u_y =0, y= ? , y-0 =0,a_y =9.8 ms^(-2)`,
` t=3s, v_y =?`
As, ` y=y_0 + u_y t+ 1/2 a_y t^2`
` =0 + 0 xx 3 + 1/2 xx 9.8 xx 3^2 = 44.1 m`
` v_y = u-y +a_y t =0 + 9.8 xx 3 = 29 .4 ms^(-10`
Let `theta` be the angle wich the resultant velocity o body while striking on ground makes with X-axis, then ` v_y /v_x= tan theta = tan 45^2 =1`
or ` v_y = u_x = 29.4ms^(-1)` .
Hence intial horizontal velocity of projection,
` u=v_x = 29.4 ms^(-1)`.