Correct Answer - `15 m s^-1`
Let `vec v_c = v_0 hat i, vec v_(h//c) = -(v_1)/(2) hat j - v_1 (sqrt(3))/(2) hat i`
`vec v_h = -(v_1)/(2) hat j + (v_0 - v_1 (sqrt(3))/(2)) hat i`
Now `v_0 - v_1 (sqrt(3))/(2) = 0 rArr v_1 = (2 v_0)/(sqrt(3))`
Also `((v_1)/(2))^2 + (v_0 - v_1 (sqrt(3))/(2))^2 = (5 sqrt(3))^2`
`rArr v_1^2 = 75 xx 4 rArr (4 v_0^2)/(2) = 75 xx 4 rArr v_0 = 15 ms^-1`.

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