Shots are fired simultaneously from the top and the bottom of a vertical cliff with elevation `37^(@)` and `53^(@)` respectively, strike an object at the ground simultaneously.If the horizontal distance of the object from the cliff is `60 m`, find the height of the cliff.

Explanation

Let `h`: height of the cliff
`T`: time of flight

For shot 2: `R = (u^(2))/(g) sin 2 theta = (u^(2))/(g)(2 sin theta cos theta)`
`60 = (u_(2)^(2))/(10)(2 sin 53^(@) cos 53^(@))`
`= (u_(2)^(2))/(10)(2 xx (4)/(5) xx (3)/(5))`
`u_(2) = 25 m//s`
`T = (2u_(2) sin theta)/(g) = (2 xx 25 sin 53^(@))/(10) = 5 xx (4)/(5) = 4 s`
For shot 1: `x = u_(1) cos 37^(@)T`
`60 = u_(1) xx (4)/(5) xx 4 rArr u_(1) = (75)/(4) m//s`
`-h = u_(1) sin 37^(@)T - (1)/(2)g T^(2)`
`= (75)/(4).(3)/(5).4 - (1)/(2).10(4)^(2) = 45 - 80`
`h = 35 m`