A ball is thrown at angle `theta` and another ball is thrown at angle `(90^(@)-theta)` with the horizontal direction from the same point with velocity `40ms^(-1)`. The second ball reaches 50m higher than the first ball. Find their individual heights. `g=10ms^(-2)`.

Explanation

For the first ball, angle of projection `= theta`, velocity of projection, `u= 40 m//s`.
Let h be the maximum height attained by it.
As maximum height attained = `(u^(2) sin ^(2)theta )/(2g)`
`therefore h (( 40 )^(2) sin^(2)theta)/( 2xx 10)" "...(1)`
For second ball, Angle of projection `= (90^(@) - theta)`.
velocity of projection, `u = 40 m//s`
Maximum height reached =` (h+50)` m
`therefore h + 50 = (u^(2) sin ^(2) (90^(@) - theta))/(2g) = ((40)^(2) cos^(2) theta)/(2xx 10) .......(2)`
By adding (1) and (2),
`2h + 50 = ((40)^(2))/(2 xx 10) xx ( sin^(2) theta + cos^(2) theta ) = ((40)^(2))/( 2xx 10) = 80`
`rArr 2h = 80 - 50 =30 rArr h = 15 m`
Height of the first ball, h = 15m & Height of the
second ball = `h+ 50 = 15 + 50 = 65 `m