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## Homework Statement

Protons are projected with an initial speed of v0 = 10^5 m/s into a region in which a uniform electric field E = 800N/C [down] is present. The protons are to hit a target that lies a horizontal distance of 10.0 mm from the point at which the protons are launched. a) find the two projection angles theta that will result in a hit b) what is the total duration of flight for each of the two trajectories?

## Homework Equations

(1) x = v0 cos (theta) t

(2) y = v0 sin (theta) t - 1/2at^2

F=qE=ma

a=(qE)/m

## The Attempt at a Solution

**x = v0 cos(theta) t y = v0 sin(theta) t - 1/2 a t^2**

when the proton hits its mark at 10 mm, the y displacement will be zero

1/2*a*t = v0 sin(theta) ===> t = (2*v0 sin(theta))/a

plugging t into the formula for the x displacement gives

x = v0 cos(theta)*(2*v0 sin(theta))/a

when the proton hits its mark at 10 mm, the y displacement will be zero

1/2*a*t = v0 sin(theta) ===> t = (2*v0 sin(theta))/a

plugging t into the formula for the x displacement gives

x = v0 cos(theta)*(2*v0 sin(theta))/a

=> x = [(v0)^2 * sin(2theta)]/a

=> theta = 0.5 *arcsin( (a*x)/(v0)^2)

=> theta = 2.2 degrees

t = (2*v0 sin(theta))/a

=> t = 10^-7 s

Now I don't really know how I am supposed to get the second angle