Question

An angle is thrown at an angle and another ball is thrown an angle (90-0) with the horizontal direction from the same point with velocity of 39.2 m/s . the second ball reaches 50 m higher than the 1st ball .find their individual hight.

Answer 1

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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