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

Explanation

`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`
`Delta t = (2u_(y))/(10) = (2 xx 10)/(10) = 2s`