Protons are projected with an initial speed of v0 = 9550 m/s into a region in which a uniform electric field E = 720N/C [down] is present. The protons are to hit a target that lies a horizontal distance of 1.27mm from the point at which the protons are launched.(adsbygoogle = window.adsbygoogle || []).push({});

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?

a) equations for projectile motion

(1) x = v0 cos (theta) t

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

gelec = |field| / mass = |qfield x E| / m

x = (R, 0) where R = 1.27mm = 1.27 x 10^-3 m

solve equation 2 for t. t = 0 or t = 2v0sin theta / g (we use this one)

(3) R = 2Vo^2 sin (theta) cos (theta) / gelec

(4) sin (theta) cos (theta) = A = a number

square equation (4) : sin (theta)^2 cos (theta)^2 = A^2

sin (theta)^2 (1 - sin (theta)^2) = A^2

sin (theta)^4 - sin (theta)^2 + A^2 = 0

using quadratic equation ... Let x be sin (theta)^2

x^2 - x + A^2 = 0

1 +/- (1 - 4A^2)^1/2 <--- equation 5

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2

a) theta = inverse sin (sqrt x)

= 53 or 37 degrees.

I'd really appreciate it if someone could check my work / answer over for me. Thanks in advance.

b) t = (2vosintheta) / g

= 2.2 x 10^-7 s

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# Projectile motion of protons

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