Q19 / 46 · NTA Abhyas (JEE Main) 2020 · Physics · Motion In One Dimension

A man in a balloon throws a stone downwards with a speed of 5 m s-1 with respect to the balloon which is moving upwards with a constant acceleration of 5 m s-2. Find the velocity (in m s-1) of the stone relative to the man after 2 s. (Take g=10 m s-2)

Correct answer: 35

Solution

Relative velocity of stone = 5 m s-1

relative acceleration of stone = 10 + 5 = 15 m s-2

  v = u + at = 5 + 15 × 2 = 35 m s-1 

 relative velocity after t = 2 s is
35 m s-1 ​ 

Hence, the correct answer is 35.

More Motion In One Dimension Questions