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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 |
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Answer» of FIRST BALL (H₁) = u²sin²Ө/2gheight of 2ND ball (H₂) =u²cos²Ө/2ga/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=0.82sin²Ө=0.18H₁ = (40)² x (0.18)/19.6 =15 mH₂ = (40)² x (0.82) /19.6 =65.43 m |
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