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Q.

A golfer standing on level ground hits a ball with a velocity of u=52ms−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=10ms−2

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a

1s

b

2s

c

3s

d

4s

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=usin⁡αt−12gt2⇒ 15=52×513t−12×10t2⇒ t2−4t+3=0⇒(t−1)(t−3)=0⇒ t=1s,t=3s. Required time is 3−1=2s
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