In horizontal level ground to projectile if at any instant velocity becomes perpendicular to initial velocity then what can you say about projection angle with horizontal.

Options

A. `theta=45^(@)`
B. `thetage45^(@)`
C. `thetale45^(@)`
D. for any value of `theta` it is possible

Answer

b

Explanation

Velocity at any time `vecv=vecu+vec(g)t`
`implies vecv=u cos theta hati+(u sin theta-gt)hatj`
Let the any time this velocity becomes perpendicular to initial velocity. The `vecv.vecu=0`
Solve to get `t=u/(g sin theta)`
Now t should be less then/equal to time of flight.
So, `tgeT`.
`u/(g sin theta)le(2u sin theta)/g implies sin^(2) thetage1/2`
`implies sin thetage1/(sqrt(2)) implies thetage45^(@)`