A particle is projected at an angle of `60^(@)` with the horizontal from the ground with a velocity `10sqrt3ms^(-1)`. The angle between velocity vector after 2s and initial velocity vector is (`g=10ms^(_2)`)

Options

A. `0^(@)`
B. `30^(@)`
C. `60^(@)`
D. `90^(@)`

Answer

D

Explanation

Initial velocity, `v_(i)=2costhetahati+4sinthetahatj=5sqrt3hati+15hatj`
Final velocity vector (after 2s),
`v_(f)=ucosthetahati+(usintheta-"gt")hatj=5sqrt3hati+5hatj`
Now, `v_(i)*v_(f)=25xx3-15xx5=0`
`thereforev_(i)botv_(f)`