A particle is projected with a speed v and an angle `theta` to the horizontal. After a time t, the magnitude of the instantaneous velocity is equal to the magnitude of the average velocity from 0 to t. Find t.

Answer

`4/3 (v sin theta)/g`

Explanation

`vecv(0)=vcosthetahati+vsinthetahatj`
`vec(v)(t)=vcostheta hati+(sintheta-g t)hatj`
|hatv(t)|=sqrt(v^(2)cos^(2)0+(vsintheta-g t)^(2))`
`ltvecv(t)=(vecv(t)+vecv(0))/2=cos theta hati+((2v sin 0-g t))/2hatj`
According to question `sqrt((v cos theta)^2+(v sin theta- g t)^2)`
`=sqrt((vcos theta)^2+((2v sin theta- g t)/2)^2)`
`v^2 cos^2 theta+(v sin theta- g t)^2=v^2cos^2theta+((2v sintheta- g t)/2)^2`
`v sin theta- g t= -v sin theta+( g t)/2implies (3 g t)/2=2v sintheta`
`t=4/3((v sin theta)/g)`