CET Karnataka Medical CET - Karnataka Medical Solved Paper-2001

  • question_answer
    A balloon starts rising from the ground with an acceleration of \[1.25\text{ }m/{{s}^{2}}\]after 8s, a stone is released from the balloon. The stone will\[(g=10m/{{s}^{2}})\]:

    A)  reach the ground in 4 second

    B)  begin to move down after being released

    C)  have a displacement of 50 m

    D)  cover a distance of 40 m in reaching the ground

    Correct Answer: A

    Solution :

     Acceleration \[a=1-25\text{ }m/{{s}^{2}}\] time t = 8 sec         (given) Acceleration due to gravity \[g=10m/{{s}^{2}}\] Let the balloon be at a height h above the ground, when the stone is released from it. \[h=0\times t+\frac{1}{2}\times 1.25\times {{(8)}^{2}}\]\[=40m\] Velocity of the balloon at this instant \[\upsilon =0+1-25\times 8=10\text{ }m/s\] For the journey of stone, let it take time t to reach the ground Now \[u=-10m/s,s=40m,g=10m/{{s}^{2}}\] so    \[s=ut+\frac{1}{2}g{{t}^{2}}\] \[40=-10t+\frac{1}{2}\times 10{{t}^{2}}\] \[{{t}^{2}}-2t-8=0\] \[(t+2)(t-4)=0\] t = - 2 (not admissible) t = 4 sec The stone takes 4 second to reach the ground.


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