Two objects are located at height 10 m above the ground. At some point of time, the objects are thrown with initial velocity `2sqrt(2)` m/s at an angle `45^(@)` and `135^(@)` with the positive X-axis, respectively. Assuming `g = 10 m/s^(2)`, the velocity vectors will be perpendicular to each other at time is equal to  

Options

A. 0.2 s
B. 0.4 s
C. 0.6 s
D. 0.8 s

Answer

B

Explanation

In this projectile motion, both objects are projected at the angle difference of `90^(@)`. They will be again have same angle difference, where they are again at same height (of 10 m) from ground. Taking, plane `O_(2)OO_(1)` as reference, time taken to teach at point `O_(1)` by first object is
Time of flight, `T_(1)=(2u sin theta)/(g)`
`=(2xx2sqrt(2)sin45^(@))/(10)=0.4" s"`
Time taken by second object to reach at point `O_(2)` is
`T_(2)=(2xx2sqrt(2)sin135^(@))/(10)=0.4" s"`
`because" "T_(1)=T_(2)`
So, velocities of both objects will be perpendicular to each other after 0.4 s.