InterviewSolution
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InterviewSolution
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The velocity of a projectile when it is at the greatest height is `(sqrt (2//5))` times its velocity when it is at half of its greatest height. Determine its angle of projection. |
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Answer» Max. height, ` H= (u^2 sin^2 theta)/(2 g)` or ` g H = ( u^2 sin^2 theta)/2` ltbRgt Velocity at highest point, ` v_h =u cos theta` Let ` v_x, v_y` be the horizontal and vetical velocity of projectile at height ` H//@`. The ltbRgt ` v_x = u cos theta` and ` v_y^2 = u^2 sin^2 theta-2 g xx H//2` ltbRgt ` = u^2 sin^@ thta-g H` ` = u^2 sin^2 theta - (u^2 sin^2 theta)/2 = (u^@ sin^2 theta)/2` Effective velocity at height H//2` `=(v_x^2 +v_y^2)^(1//2)` ltbRgt As per queston, ` sqrt 2/5 (v_x^2 +v_y^2)^(1//2) = v_h ` or ` 2/5 (v_x^2+v_y^2) =v_h^2` or ` sin^2 theta = 3 cos^2 theta` or ` tan theta =sqrt 3 cos theta` or ` tan theta = sqrt 3 = tan 60^2 ` or ` theta = 60^@`. |
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