InterviewSolution
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InterviewSolution
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The trajectory of a projectile in a vertical plane is `y = ax - bx^2`, where `a and b` are constant and `x and y` are, respectively, horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projectile from the horizontal are.A. `(b^2)/(2 a), tan^-1 (b)`B. `(a^2)/(b), tan^-1 (2b)`C. `(a^2)/(4 b), tan^-1 (a)`D. `(2 a^2)/(b), tan^-1 (a)` |
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Answer» Correct Answer - C ( c) `y = ax - bx^2,` for height of `y` to be maximum : `(d y)/(d x) = 0` or `a - 2 bx = 0` or `x = (a)/(2 b)` (i) `y_(max) = a((a)/(2 b)) - b ((a)/(2 b))^2 = (a^2)/(4 b)` (ii) `((d y)/(d x))_(x = 0) = a = tan theta_0`, where `theta_0` = angle of projection `theta_0 = tan^-1 (a)`. |
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