Question

a ball is thrown at angle theta and another ball is thrown at angle 90-theta with the horizontal from the same point with velocity 40 m/s

Answer 1

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 = (40)² x (0.18)/19.6

    =15 m

H₂ = (40)² x (0.82) /19.6

    =65.43 m

Answer 2

Explanation:

Answer refer in Attachment

Thank you