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A golfer standing on the ground hits a b...

A golfer standing on the ground hits a ball with a velocity of 52 mis at an angle Q above the 5 horizontal if tan `theta = 5/12` find the time for which the ball is LB at least 15m above the ground? (g = `m//s^(2)`)

Text Solution

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`v_(y) = sqrt(u_(y)^(2) -2gy)`
`=sqrt(52 xx 52 xx (5 xx 5)/(13 xx 13) -2 xx 10 xx 15)`
`= sqrt(16 xx 25 -300) = 10`
`Deltat =(2u_(y))/10 =(2 xx 10)/10 = 2s`
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