9th Class Science Time and Motion Question Bank Motion

  • question_answer
    A ball is dropped on the floor from a height of 10 m. It rebounds to a height of 2.5 m. If the ball is in contact with the floor for 0.01 sec, then average acceleration during contact is

    A)  \[\text{21}00\text{m}/{{\text{s}}^{\text{2}}}\]  

    B)  \[\text{14}00\text{ m}/{{\text{s}}^{\text{2}}}\]

    C)  \[\text{7}00\text{ m}/{{\text{s}}^{\text{2}}}\]

    D) \[~\text{4}00\text{ m}/{{\text{s}}^{\text{2}}}\]

    Correct Answer: A

    Solution :

     Let u be the velocity with which the ball hits the ground, then          \[{{u}^{2}}=2gh\]             \[=2\times 9.8\times 10=196\]   \[\therefore \]    \[u=14\,\,m/\sec \] If \[v\] be the velocity with which it rebounds, then             \[{{v}^{2}}=2\times 9.8\times 2.5=49\] \[\Rightarrow \]   \[v=7\,\,m/\sec \] \[\therefore \]      \[\Delta v=(v-u)\]             \[=7(m/\sec )-(-14m/\sec )\]             \[=21\,\,m/\sec \] \[\therefore \]      \[a=\frac{\Delta v}{\Delta t}=\frac{21}{0.01}=-2100\,\,m/{{s}^{2}}\](upward)


You need to login to perform this action.
You will be redirected in 3 sec spinner