Question

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 $m/s^2$)

• A. 1 sec
• B. 1.414 sec
• C. 2.828 sec
• D. 2 sec

Answer 1

The correct option is B 1.414 sec

$t=\sqrt{\frac{2h}{g+a}}=\sqrt{\frac{2h}{g+\frac{Eq}{m}}}$
$t=\sqrt{\frac{2\times 20}{10+\left(\frac{{10}^{-2}\times 2}{2\times {10}^{-3}}\right)}}=\sqrt{\frac{40}{10+10}}=\sqrt{\frac{40}{20}}$
$t=\sqrt{2}=1.414sec$