A ball is thrown at an angle ` theta` and another ball is thrown at an angle ` (90^@- tehta) with the horizontal direction from the same point, each with velocity ` 40 ms^(-1)`. The second ball reaches ` 40 m` higher than the first ball. Find their individul heithts, Take ` g= 10 m//s`.

Explanation

For the first ball, angle of projection ` = tehta`,
` u= 40 ms^(-10`
Maximum height,
` H_1 = ( u^2 sin^2 theta)/( 2g) = (( 40)^2 sin^2 theta)/ (2 xx 10)`
For second ball. Angle of projection
` = (0^@- theta), u= 40 ms^(-1)`
Maximum height,
` H_2 = H_1 + 40 = (u^2 sin^2(90^@- theta)/ (2g)
` = ((40)^2 cos^2 theta)/(2 xx 10)`
Adding (i) and (ii)
` 2 H_1 + 40 = (40)^2/(20) [sin^2 theta+ cos^@ theta] =80`
or ` 2H_1 80- 40 =40` or ` H_1 = 20 m`
H_2 = H_1 = 40= 20 = 40 = 60 m` .