CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2001

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

    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 \[a=1.25\,m/{{s}^{2}}\] time\[t=8\,sec\]                                   (given) Acceleration due to gravity \[g=10\,m/{{s}^{2}}\] Let the balloon be at a height h above the ground, when the stone is released from it. \[h=0\times t+\frac{1}{2}\times 1.25\,\times {{(8)}^{2}}\] \[=40\,m\] Velocity of the balloon at this instant \[\upsilon =0+1.25\times 8=10\,m/s\] For the journey of stone, let it take time t to reach the ground Now \[u=-10\,m/s,\]\[s=40m,\,g=10\,m/{{s}^{2}}\] 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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