Question

Protons are projected with an initial speed `v_i = 9.55 xx 10^3 m//s` into a region where a uniform electric field `E = (-720hat j`) N/C is present, as shown in figure. The protons are to hit a target that lies at a horizontal distance of 1.27 mm from the point where the protons are launched. Find
(a) the two projection angles 0 that result in a hit and
(b) the total time of flight for each trajectory.

Answer 1

Correct Answer - A::B
a. `g_e=g+(qE)/m`
`=10+((1.6xx10^-19)(720))/(1.67xx10^-27)`
`~~6.9xx10^10m//s^2`
Now, `R=(u^2sin2theta)/g_e`
`:. (1.27xx10^-3)=((9.55xx10^3)^2sin2theta)/(6.9xx10^10)`
Solving this equation we get
`theta=37^@ and 53^@`
b. Applying `T=(2usintheta)/g_e`