question_answer

A ball is thrown vertically upwards with a speed of 10 m/s from the top of the tower 200m high and another is thrown vertically downwards with the same speed simultaneously. The time difference between them in reaching the ground is (\[\text{g}=\text{1}0\text{ m}/{{\text{s}}^{\text{2}}}\])

A)  12s                       

B)  6s   

C)  2s                          

D)  1s

Correct Answer: C

Solution :

  \[U=10m/s\] Time taken to the ball thrown downwards to reach the surface is \[\frac{u}{g}={{T}_{1}}\] And time taken to the ball thrown up to reach the max. height and come to point of projection is\[\frac{2u}{g}\]and to reach the ground it takes u / g . Total time\[\frac{u}{g}+\frac{2u}{g}={{T}_{2}}.\] \[{{T}_{2}}-{{T}_{1}}=\frac{u}{g}+\frac{2u}{g}-\frac{u}{g}=\frac{2\times 10}{10}=2s\]