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The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of the motion of the projectile, giving it a constant horizontal acceleration = g/2. Under the same conditions of projection, the horizontal range of the projectile will now be(a) R + H/2.(b) R + H (c) R + 3H/2 (d) R + 2H |
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Answer» Correct Answer is: (d) R + 2H For the projectile, R = u2 sin 2θ/g, H = u2 sin2 θ/ 2g Time of flight = t = 2u sin θ/g. Initial horizontal velocity = u cos θ. When horizontal acceleration g/2 is present, the horizontal range = (u cos θ)t + 1/2 (g/2) t = (u cos θ) 2u sin θ/g + g/4 (4u2 sin2 θ/g2) = R + 2H. |
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