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)`.A. 1 sB. 2 sC. 3 sD. 4 s

A golfer standing on level ground hits a ball with a velocity of `52 m

1.	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)`.A. 1 sB. 2 sC. 3 sD. 4 s
Answer» Correct Answer - B (b) Let at any time `t`, the ball be at height of `15 m`. `S_y = u_y t + (1)/(2) a_y t^2` `rArr 15 = u sin prop t - (1)/(2) "gt"^2` `rArr 15 = 52 xx (5)/(13) t - (1)/(2) xx 10 t^2` `rArr t^2 - 4t + 3 = 0 rArr (t -1)(t - 3) = 0` `rArr t = 1 s, t = 3 s`. Required time is `3 - 1 = 2 s`.

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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)`.A. 1 sB. 2 sC. 3 sD. 4 s

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