Manipal Engineering Manipal Engineering Solved Paper-2010

  • question_answer
    A ball rolls off the top of stairway with a horizontal velocity of magnitude 1.8 m/s. The steps are 0.20 m high and 0.20 m wide. Which step will the ball hit first?

    A)  First                                     

    B)  Second

    C)  Third                    

    D)         Fourth

    Correct Answer: D

    Solution :

    Given,\[x=0.20m,\,\,y=0.20m,\,\,u=1.8m/s\]. Let the ball strikes the nth step of stairs, Vertical distance travelled                 \[=ny=n\times 0.20=\frac{1}{2}g{{t}^{2}}\] Horizontal distance travelled,\[nx=ut\] or            \[t=\frac{nx}{u}\] \[\therefore \]  \[ny=\frac{1}{2}g\times \frac{{{n}^{2}}{{x}^{2}}}{{{u}^{2}}}\] or            \[n=\frac{2{{u}^{2}}}{g}\frac{y}{{{x}^{2}}}=\frac{2\times {{(1.8)}^{2}}\times 0.20}{9.8\times {{(0.20)}^{2}}}=3.3\approx 4\]


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