1.

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))`

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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