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EEJ MAIN Physics QUESTION #7096
Question 1
A ball dropped from $5\text{ m}$ bounces back to $81/100$ of its previous height repeatedly. Find its average speed ($g=10\text{ ms}^{-2}$).
  • $3.50\text{ ms}^{-1}$
  • $2.0\text{ ms}^{-1}$
  • $2.50\text{ ms}^{-1}$✔️
  • $3.0\text{ ms}^{-1}$
Correct Answer Explanation

Average speed = $\frac{\text{Total Distance}}{\text{Total Time}}$.

  • Total Distance $D = H + 2e^2H + 2e^4H... = H(\frac{1+e^2}{1-e^2})$ where $e^2 = 0.81$.
  • Total Time $T = \sqrt{\frac{2H}{g}} + 2e\sqrt{\frac{2H}{g}}... = \sqrt{\frac{2H}{g}}(\frac{1+e}{1-e})$.
  • With $H=5, g=10, e=0.9$, average speed results in $2.50\text{ ms}^{-1}$.