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The horizontal range of a projectile is `2 sqrt(3)` times its maximum height. Find the angle of projection. |
Answer» If `u and prop` are the initial velocity of projection and angle of projection, respectively, then Maximum height attained `=(u^2 sin^2 prop)/(2 g)` Horizontal range `(2 u^2 sin prop cos prop)/(g)` According to the problem, `(2 u^2 sin prop cos prop)/(g) = 2 sqrt(3) ((u^2 sin^2 prop)/(2 g)) rArr tan prop = ((2)/(sqrt(3)))` `rArr prop = tan^-1 ((2)/(sqrt(3)))`. |
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