- #1
Elvis 123456789
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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
=> 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