A stone is projected from level ground at `t=0` sec such that its horizontal and vertical components of initial velocity are `10(m)/(s)` and `20(m)/(s)` respectively. Then the instant of time at which tangential and normal components of acceleration of stone are same is: (neglect air resistance)`g=10(m)/(s^2)`.

Options

A. `(1)/(2) sec`
B. `(3)/(2) sec`
C. 3 sec
D. 4 sec

Answer

C

Explanation

Tangential acceleration =`a_(t)=g sin theta`
Normal acceleration =`a_(n)=g cos theta`
`a_(t)=a_(n)`
`g sin theta=g cos thetarArr theta=45^(@)`
`rArr v_(y)=v_(x)`
`u_(y)-g t=u_(x)`
`20-(10)t=10`
`t=1 sec`
During downward motion
`a_(t)=a_(n)`
`v_(y)=-v_(x)`
`20-10t=-10`
`rArr t=3 sec`