A golfer standing on level ground hits a ball with a velocity of u = 52 ms-1 at an angle α above the horizontal. If tan α = 5/12, then the time for which the ball is at least 15 m above the ground will be (take g = 10 ms-2)

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1 s

2 s

3 s

4 s

answer is B.

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Detailed Solution

Let at any time t, the ball be at height of 15 m.Sy=uyt+12ayt2 15=usinαt-12gt2 15=52×513t-12×10t2 t2-4t+3=0 (t - 1)(t - 3) = 0 t = 1 s, t = 3 s. Required time is 3 - 1 = 2 s

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