A ball is thrown from the top of a building `45 m` high with a speed `20 m s^-1` above the horizontal at an angle of `30^@`. Find
(a) The time taken by the ball to reach the ground.
(b) The speed of ball just before it touches the ground.

Explanation

Given `v = 20 m s^-1 , theta = 30^@, H = 45 m`.
(a) As the ball has been projected at an angle of `30^@` above horizontally, so first of all we need to analyse the velocity horizontally and vertically. This will be useful while using distance - time relation in horizontal and vertical directions.
`v_(x i) = v cos 30^@ = 20 xx (sqrt(3))/(2) = 10 sqrt(3) m s^-1`
`v_(y i) = v sin 30^@ = 20 xx (1)/(2) = 10 m s^-1`
It will be easy for us to use distance - time relation in vertical as it will involve less calculation.
In y - direction : `-45 = 10 t - (1)/(2) xx "gt"^2 rArr t^2 - 2 t = 0`
which on solving gives `t = 1 + sqrt(10)) s` (positive value), (other value is `1 - sqrt(10)) s`, a negative value of time is not acceptable).
(b) `v_(y f) = 10 - 10 xx (1 + sqrt(10))) = -10 sqrt(10)) m s^-1`
`v_f = sqrt(v_(y f)^2 + v_(x f)^2) = sqrt((10sqrt(10))^2 + (10 sqrt(3)^2)`
=`10 sqrt(3)) m s^-1`
.