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

(i) Let H be the height of the tower
The time of length, `T_(f) = sqrt((2H)/(g)) = 3s`
`rarr H = (g xx (3)^(2))/(2) = (9.8 xx 9)/(2) = 44.1 m`
(ii) Let the speed of projection be `v_(0)`.
Then for horizontal projection
`v_(x) = v_(0) rArr v_(y) =- g t`
At `t = T_(f) = 3s, v_(y) = - 9.8 xx 3 =- 29.4 ms^(-1)`
The angle which the final velocity makes with the horizontal `= theta = 45^(@)` (Given)
`rArr - tan 45^(@) = (-v_(y))/(v_(x)) rArr v_(y) = v_(x)`
So, `v_(x) = 29.4 ms^(-1)`