Question

A particle of mass 2 Kg and charge 1 mC is projected vertically with a velocity $10m{s}^{-1}.$ There is a uniform horizontal eletric field of ${10}^{4}N/C.$. Then:

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

Answer 1

Answer: (A), (B), (C)

The maximum height reached is 5 m

Time of flight $\left(t\right)=\frac{2u}{g}=\frac{2\times 10}{10}=2sec$
$H=\frac{u^2}{2g}=\frac{10\times 10}{2\times 10}=5m$
$R=0+\frac{1}{2}\left(\frac{qE}{m}\right)t^2=\frac{1}{2}\times \frac{{10}^{-3}\times {10}^4\times 2\times 2}{2}=10m$

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.