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 10m
B. the time of flight of the particle is 2 s.
C. the maximum height reached is 5 m.
D. the horizontal range of the particle is 0.

Answer

A, B, C

Explanation

a.,b.,c.
Time of flight `1(t)=(2u)/(g)=(2xx10)/(10)=2s`
`H=(u^(2))/(2g)=(10^(2))/(2xx10)=5m`
`R=0+(1)/(2)((qE)/(m))t^(2)=(1)/(2)xx(10^(-3)xx10^(4)xx2xx2)/(2)=10m`