An object is launched from a cliff 20 m above the ground at an angle of `30^(@)` above the horizontal with an initial speed of 30m/s. How far horizontally does the object travel before landing on the object travel before landing on the ground ? (g ` = 10 m//s^(2)`)

Explanation

Height of cliff = 20 m
Angle of projection , `theta = 30^(@)`
Velocity of projection u = 30 m /s
Total horizontal distance travelled = OC = OB' + B'C
a) But OB' = AB = Range R R `= (u^(2) sin 2 theta)/(g)`
`:. R = (30 xx 30 sin 60^(@))/(10) = 90 (sqrt(3))/(2) = 45 sqrt(3) to (1)`
Distance B'C = Range of a horizontal projectile .
`:. ` Range = u cos `theta` t
u. cos `theta = 30 * (sqrt(3))/(2) = 15 sqrt(3)*`
Time taken to reach the ground , t = ?
Given `S_(y) = 20, u_(y) = u sin theta = 30 sin 30^(@) = 15 m// s `
`:. S_(y) = 20 = 15 t + (10)/(2) t^(2) 5t^(2) + 15 t - 20 = 0 `
or `t^(2) + 3t - 4 = 0 ` or (t + 4) (t - 1) = 0
`:.` t = - 4 or t = 1 ` :.` t is Not - ve use t = 1
`:.` Range = 4 . cos `theta * t = 15 sqrt(3) to (2)`
Total distance travelled before reading the ground ` = 45 sqrt(3) + 15 sqrt(3) = 60 sqrt(3)` m .