A ball is thrown at angle 0 and another ball is thrown at angle `(90^(@) – 0)` with the horizontal direction from the same point with velocity `40" ms"^(–1)`. The second ball reaches 50m higher than the first ball. Find their individual heights. `g = 10" ms"^(–2)`

Explanation

For the first ball, angle of projection `=theta`, velocity of projection, `u=40 ms^(-1)`
Let h be the maximum height attained by it.
As maximum height `=(u^(2)sin^(2)theta)/(2g) therefore h=((40)^(2)sin^(2)theta)/(2xx10)`....(1)
For second ball, Angle of projection `-(90^(@)-theta)`. velocity of projection, `u=40ms^(-1)`
Maximum height reached `=(h+50)m`
`therefore h+50=(u^(2)sin^(2)(90^(@)-theta))/(2g)=((40)^(2)cos^(2)theta)/(2xx10)` .....(2)
By adding (1) and (2),
`2h+50=((40)^(2))/(2xx10)xx(sin^(2)theta+cos^(2)theta)=((40)^(2))/(2xx10)=80rArr 2h=80-50=30rArrh=15m`
Height of the first ball, h=15m `&` Height of the second ball, =h+50=15+50=65m