Question

A ball is thrown at an angle theta and another ball is thrown at an angle (90-theta) with the horizontal direction from the same point wih velocity 39.2 m/s. The second ball reaches 50m higher than the first ball. Find their individual heights. bla

Answer 1

height of first ball (H1) = u^2sin^2@/2g

height of 2nd ball (H2) =u^2cos^2@/2g

a/c to question ,
H 2 - H1 = 50m
u^2cos2@/2g=50
(39.2)^2 cos2@=50 x 19.6
cos2@= 50/2 x 39.2 = 0.64
2cos^2@=1.64
cos^2@=0.82

sin^2@=0.18

now,
H1 = (39.2)^2 x (0.18)/19.6
=14.4 m

H2 = (39.2)^2 x (0.82) /19.6
=64.288 m

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