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BCECE Engineering BCECE Engineering Solved Paper-2006

  • question_answer
    A can is moving horizontally along a straight line with constant speed 30 m/s. A projectile is to be fired from the moving cart in such a way that it will return to the cart after the cart has moved 80 m. At what speed (relative to the can) must the projectile be fired? (Takes \[=10\,m/{{s}^{2}}\])

    A)  10 m/s                                

    B)         \[10\sqrt{8}\]m/s

    C)         \[\frac{40}{3}\] m/s                                      

    D)         None of these

    Correct Answer: C

    Solution :

    As seen from the cart the projectile moves vertically upwards and comes back. The time taken by cart to cover 80 m                                 \[=\frac{s}{v}=\frac{80}{30}=\frac{8}{3}s\]                 Here,\[u=?,\,v=0,a=-g=10m/{{s}^{2}}\] (for a projectile going upwards) and        \[t=\frac{8/3}{2}=\frac{4}{3}s\] From first equation of motion \[v=u+at\] \[0=u-10\times \frac{4}{3}\]             \[\therefore \]      \[u=\frac{40}{3}m/s\]


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