A ball is thrown from a point at a distance `40 m` from a wall of height `15 m`. It just clears the wall and then attains maximum height. Find the maximum height the angle of projection is `45^(@)`.

Explanation

Equation of trajectory
`y = x tan theta - (g x^(2))/(2u^(2) cos^(2) theta)`
`15 = 40 tan 45^(@) - (10(40)^(2))/(2u^(2) cos^(2) 45^(@))`
`= 40 - (16000)/(u^(2))`
`(16000)/(u^(2)) = 25 rArr u^(2) = (16000)/(25) = 640`
`H_(max) = (u^(2) sin^(2) theta)/(2g) = (640 sin^(2) 45^(@))/(20) = 16 m`
OR
`y = x tan theta(1 - (x)/( R))`
When `x = 40 m, y = 15 m, theta = 45^(@)`
`15 = 40 tan 45^(@)(1 - (40)/(R ))`
`(15)/(40) = 1 - (40)/(R ) rArr (40)/(R ) = 1 - (3)/(8) = (5)/(8) rArr R = 64 m`
`R = (u^(2))/(g) sin 2theta rArr 64 = (u^(2))/(g)sin 90^(@) rArr (u^(2))/(g) = 64`
Hence `H_(max) = (u^(2) sin^(2) theta)/(2 g) = (1)/(2) (64) sin^(2) 45^(@) = (64)/(4) = 16 m`