Motion In One Dimension Question 4
Question 4 - 29 January - Shift 1
A tennis ball is dropped on to the floor from a height of $9.8 m$. It rebounds to a height $5.0 m$. Ball comes in contact with the floor for $0.2 s$. The average acceleration during contact is $ms^{-2}$. [Given $g=10 ms^{-2}$ ]
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Answer: (120)
Solution:
Formula: Average acceleration
$ \begin{aligned} & v_i=\sqrt{2 gh_i} \\ & =\sqrt{2 \times 10 \times 9.8} \downarrow \\ & =14 m / s \downarrow \\ & v_f=\sqrt{2 gh_f} \\ & =\sqrt{2 \times 10 \times 5} \uparrow \\ & =\mathbf{1 0} \mathbf{~ m} / s \uparrow \end{aligned} $
$ |\overrightarrow{{}a}_avg^{m}|=|\frac{\Delta \overrightarrow{{}v}}{\Delta t}|=\frac{24}{0.2}=120 m / s^{2} $
