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

As shown in figure, suppose the body is thrown horizontally from the top O of a tower of height y with velocity u. The body hits the ground after 3s. Considering vertically downward motion of the body,
`y=u_yt+1/2"g"t^2=0xx3+1/2xx9.8xx(3)^2=44.1m`
[`:.` Initial vertical velocity `u_y=0`]
Final vertical velocity
`v_y=u_y+gt=0+9.8xx3=29.4ms^(-1)`
Final horizontal velocity `v_x=u`
As the resultant velocity u makes an angle of `45^(@)` with the horizontal , so
`tan45^(@)` or `1=(29.4)/x` or u`=29.4ms^(-1)`