One body is thrown at an angke ` theta` with the horizontal and another similar body is thrown at an angle `theta` with the vertical direction from the same point with same velocity` 40 ms^(-1)`. The second body reaches ` 50` metrea higher than the first body. Deteramine their individual heights. Take `g= 10 ms^(-2)`.

Explanation

For the first body, angle of projection is `theta` with the horzontal directin. Velocity of projection `u=40 m//s`
Maximum heitght,
`H_(1) =(u^(2) sin^(2) theta) /(2 g) =((40)^(2) sin ^(2) theta)/(2xx 10)` .(i)
For the second body, angle of projection ` (90^(@) -theta) with the horizontal direction, velocity of projection, `u=40 ms^(-1)`.
Maximu height reached, `H_(2) =H_(1) +50 `...(ii)
Where
`H_(2) =((40)^(2) sin^(2) (90^(@)-theta))/(2xx 10) =((40)^(2) cos^(2) theta)/(2 xx 10)`
Form (iii), wr have
:. H_(1) + 50 =((40)^(2) cos^(2) theta)/(2 xx 10) =((40)^(2) cos^(2) theta)/(2 xx 10)`
Adding (i) and (iii), we have
`2 H_(1) + 50 =((40)^(2))/((40))^(2)/(2 xx 10) [sin ^(2) theta + cos^(2) theta]`
`=((40)^(2))/(20) =80 or `2 H_(1) =80 -50 =30`
or `H_(1) =15 m`
Height of the first body, H_(1) =15 m`
Height of the second body, `H_(2) =15 + 50 =65 m`.