An electron is projected into a uniform electric field that has a magnitude of 500 N/C.
The direction of the electric field is vertically upward. The initial velocity of the
electron has a magnitude of 6.00 × 10^6 m/s, and its direction is at an angle of 30°
above the horizontal. Find:
I. The maximum distance the electron rises vertically above its initial elevation.
II. After what horizontal distance does the electron return to its original elevation.
Answer
Acceleration of electron
Time of motion
Maximum vertical distance
Horizontal distance
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