A body of mass 2 gm is projected horizontally from the top of tower of height 20m with a velocity 10 m/s. The charge on the body is 2C. Electric field is a;;lied vertically downwards and of intensity $10^{- 2}$ N/C. Find the time taken by the body to touch the ground (g = 10 $m / s^{2}$ )

1 sec

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1.414 sec

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2.828 sec

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2 sec

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Solution

The correct option is B 1.414 sec
$t = \sqrt{\frac{2 h}{g + a}} = \sqrt{\frac{2 h}{g + \frac{E q}{m}}}$
$t = \sqrt{\frac{2 \times 20}{10 + (\frac{10^{- 2} \times 2}{2 \times 10^{- 3}})}} = \sqrt{\frac{40}{10 + 10}} = \sqrt{\frac{40}{20}}$
$t = \sqrt{2} = 1.414 s e c$

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