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)`.

Options

A. 1 s
B. 2 s
C. 3 s
D. 4 s

Answer

B

Explanation

(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`.