Q.
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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a
1 s
b
2 s
c
3 s
d
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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