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Question

# A particle of mass 2 Kg and charge 1 mC is projected vertically with a velocity 10 ms−1. There is a uniform horizontal eletric field of 104 N/C.. Then:

A
The horizontal range of the particle is 10 m
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B
The time of flight of the particle is 2 s
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C
The maximum height reached is 5 m
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D
The horizontal range of the particle is 0
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Solution

## The correct option is C The maximum height reached is 5 mTime of flight (t)=2ug=2×1010=2 sec H=u22g=10×102×10=5 m R=0+12(qEm)t2=12×10−3×104×2×22=10 m Why this question? Tip: i) Time of flight and maximum height are independent of acceleration in the horizontal direction. ii) The maximum height is at R/2, only if the net acceleration is perpendicular to the range.

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