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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=10m//s^(2))` |
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Answer» Correct Answer - `(2)` `y = u sin theta t - (1)/(2) "gt"^2` `15 = 52 xx (5)/(13) t - (1)/(2) xx 10 t^2` `5t^2 + -20t + 15 = 0` `t^2 - 4t + 3 = 0` `t^2 - 3t xx t + 3 = 0 rArr t (t - 3)-1(t - 3) = 0` Thus, `t_2 - t_1 = 3 - 1 = 2 s`. |
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