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A golfer standing on level ground hits a ball with a velocity of `52 m s^-1` at an angle `theta` above the horizontal. If `tan theta = 5//12`, then find the time for which then ball is atleast `15 m` above the ground `(take g = 10 m s^-2)`. |
Answer» `y=u sin alpha.t-1/2 "gt"^(2)` | |