InterviewSolution
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InterviewSolution
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If at a height of ` 40 m`, the direction of moton of a projectile makes an angle `pi //4` with the orizontal, then its initial velocity and angle of projection are, respectively,A. ` 30 , 1/2 cos^(-1) (- 4/5)`B. ` 30 1/2 cos^(-1) (-1/2)`C. ` 50 1/2 cos^(1) (- 8 /(25)`D. 60 1/2 cos^(-1) (- 1/4)` |
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Answer» Correct Answer - C Let (u) be the initial velocity of projectile and ` theta` be the angel of projmection. Taking vertical upwarawd motion for hieght (h) , we have ` u =u sin theta, a=- g, S=h, v=v-y, As ` v^2 = u^2 + 2 aS, so ` v_y^2 = u^2 sin^2 theta- 2 gh` ...(i) At height ` h, tahn (pi)/4 = v_y/v_x ` or ` v_y =v-x = u cos theta` :. ` u^2 cos^2 theta = u ^2 sin^2 sin^2 theta - 2 gt` or ` u^2 (cos^2 theta- sin ^2 theta)=- 2 gt` ltBrgt or ` u^2 cos 2 theta =- 2 gt ` ....(ii) The equation (ii) is valid only is ` u= 50 m//s` and ` cos ` 2 theta =- 8//25` or ` theta 1/2 cos ^(-1) (- 8//25)`. |
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