NEET AIPMT SOLVED PAPER SCREENING 2010

  • question_answer
    A ball is dropped 'from a high rise platform at t = 0 starting from rest. After 6 s another ball is thrown downwards from the same platform with a speed v. The two balls meet at t = 18 s. What is the value of v ? (take \[g=10\,m{{s}^{-2}}\])

    A) \[\text{74}\,\text{m}{{\text{s}}^{-2}}\]                

    B) \[55\,\text{m}{{\text{s}}^{-1}}\]              

    C)        \[40\,\text{m}{{\text{s}}^{-1}}\]              

    D)        \[60\,\text{m}{{\text{s}}^{-1}}\]

    Correct Answer: A

    Solution :

    For first ball, u = 0 \[\therefore \]  \[{{s}_{1}}=\frac{1}{2}gt_{1}^{2}=\frac{1}{2}\times g{{(18)}^{2}}\] For second ball, initial velocity = v \[\therefore \]  \[{{s}_{2}}=v{{t}_{2}}+\frac{1}{2}g{{t}^{2}}\]                 \[{{t}_{2}}=18-6=12s\] \[\Rightarrow \]               \[{{s}_{2}}=v\times 12+\frac{1}{2}g{{(12)}^{2}}\] Here,     \[{{s}_{1}}={{s}_{2}}\]                 \[\frac{1}{2}g{{(18)}^{2}}=12v+\frac{1}{2}g{{(12)}^{2}}\] \[\Rightarrow \]               \[v=74\,\,m{{s}^{-1}}\]


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