At a certain height a body at re4stt explodes into two equal fragments with one fragments receiving a horizontal velocity 10 m/s.The time interval after the explosion for which the velocity vectors of the two fragments become perpendicular to each other is (g = 10 m/s2)Please help me out with the solution of this question in physics.I am Shanmukhi Sai Sindhuja from Hyderabad

The time interval after which both the velocity vector will be perpendicular to each other will be 1 sec
Explanation: 
From the law of conservation of momentum
m x 0 = mv/2 + mu/2
=> v = -u 
where u and v are velocity of both the fragments.
Given u = 10 m/s
=> |v| = |u| = 10 m/s
If Ф is the angle between both the velocity component, then
t = (√uv)/g x cotФ/2
given,
u = v = 10 m/s
g = 10 m/s²
Ф = 90 => cotФ/2 = cot 45 = 1
hence 
t = (√10 x 10)/10 x 1
=> t = 10/10 = 1 sec
Hence the time interval after which both the velocity vector will be perpendicular to each other will be 1 sec