Question

A particle is projected with a velocity `vecv=8hati+6hatj m//s`. The time after which it will starts moving perpendicular to its initial direction of motions is
A. `0.5s`
B. `1.25s`
C. `1s`
D. `5//3s`

Answer 1

Correct Answer - d
Given `vecv_(1)=8hati+6hatjm//s^(2)`
Acceleration due to gravity, `veca=-ghati=-10hatjm//s^(2)`
Velocity after time t, `vecv_(2)=vecv_(1)+vec(a)t`
`vecv_(2)=(8hati+6hatj)-10thatj=8hatj+(6-10t)hatj`
Now `vecv_(1) and vecv_(2)` will be perpendicular if `vecv_(1).vecv_(2)=0`
`(8hati+6hatj).[8hati+(6-10t)hatj]=0`
`64+6(6-10t)=0`
`64+36-60t=0`
`60t=100t=5/3 sec`