Question

A ball is thrown at an angle theta and another ball is thrown at an angle of 90-theta with the horizontal direction from the same point with velocity 39.2 metre per second the second ball reaches 50m higher than the first ball find individual height.

Answer 1

We can see that the question mentions a ball thrown with an angle with the horizontal direction, so from this we could figure our that the question should be solved in context to the chapter: motion in a plane. Then its only a matter of time to figure out which equation to use....

Answer 2

height of first ball (H₁) = u²sin²Ө/2g

height of 2nd ball (H₂) =u²cos²Ө/2g

a/c to question ,

H₂ - H₁ = 50m

⇒u²cos2Ө/2g=50

⇒(39.2)² cos2Ө=50 x 19.6

⇒cos2Ө= 50/2 x 39.2 = 0.64

⇒2cos²Ө=1.64

⇒cos²Ө=0.82

sin²Ө=0.18

H₁ = (39.2)² x (0.18)/19.6

    =14.4 m

H₂ = (39.2)² x (0.82) /19.6

    =64.288 m