2. Solved Papers
  3. NEET

NEET AIPMT SOLVED PAPER 2003

  • question_answer
                    A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time?                 (Given \[g\]\[=9.8\,m/{{s}^{2}}\])

    A)                                                                                                                                                            Any speed less than 19.6 m/s                    

    B)                 Only with speed 19.6 m/s

    C)                 More than 19.6 m/s       

    D)                 At least 9.8 m/s

    Correct Answer: C

    Solution :

                                                           Time taken by ball to reach maximum height                 \[v=u-gT\]                 at maximum height, final speed is zero i.e.,                 v = 0 So,          u = gT or            T = u/g In            2 s, \[u\,=2\times 9.8=19.6\,\,m/s\]                 If man throws the ball with velocity of 19.6 m/ s then after 2 sec it will reach the maximum height. When he throws 2nd ball, 1st is at top. When he throws third ball, 1st will come to ground and 2nd will at the top.                 Therefore, only 2 balls are in air. If he wants to keep more than 2 balls in air he should throw the ball with a speed greater than 19.6 m/s.


You need to login to perform this action.
You will be redirected in 3 sec spinner