INT A proton is fired horizontally into a 1.0×10^5 N/C vertical electric field. It rises 1.0 cm vertically after having traveled 5.0 cm horizontally. What was the proton's initial speed?
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Identify the known values: Electric field strength (E) = 1.0 \times 10^5 \, \text{N/C}, vertical displacement (y) = 1.0 \, \text{cm} = 0.01 \, \text{m}, horizontal displacement (x) = 5.0 \, \text{cm} = 0.05 \, \text{m}, and the charge (q) and mass (m) of a proton (q = 1.602 \times 10^{-19} \, \text{C}, m = 1.672 \times 10^{-27} \, \text{kg}).
Calculate the vertical acceleration of the proton due to the electric field using the formula: a = \frac{qE}{m}.
Use the kinematic equation for vertical motion to find the time (t) it takes for the proton to travel vertically: y = \frac{1}{2}at^2. Solve for t.
Assuming no horizontal acceleration, use the horizontal displacement and the time calculated in the previous step to find the initial horizontal speed (v_x) of the proton using the formula: x = v_x \cdot t.
The initial speed of the proton is the magnitude of the horizontal velocity component, which is the value calculated for v_x.