A particle of mass 2 kg chrge 1 mC is projected vertially with velocity k`10 ms^-1`. There is as uniform horizontal electric field of `10^4N//C,` then

Options

A. the horizontal range of the particle is `10 m`
B. the time of flight of the particle is `2s`
C. the maximum heighty reached is `5m`
D. the horizontal range of the particle is `5m`

Answer

A, B, C

Explanation

`T=(2u_y)/g=(2xx10)/10=2s`
`H=u_y^2/(2g)=((10)^2)/20=5m`
`R=1/2a_xT^2=1/2((qE)/m)T^2`
`=1/2((10^-3xx10^4)/2)(2)^2`
`=10m`