Question

A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to $\frac{81}{100}$ of the height through which it falls. Find the average speed of the ball.

(Take g = 10 ms$^{-2}$)

• A. 3.0 ms$^{-1}$
• B. 3.50 ms$^{-1}$
• C. 2.0 ms$^{-1}$
• D. 2.50 ms$^{-1}$

Answer 1

Answer: (D)

2.50 ms$^{-1}$

Answer 2

The average speed of the ball can be found by dividing the total distance traveled by the total time of motion.

1. Initial Drop:

   • Height $h = 5$ m.
   • Time for the first fall: $$t_0 = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5}{10}} = 1 \text{ s}$$

2. Bounces:

   • Each bounce rises to a fraction $r = \frac{81}{100} = 0.81$ of the previous fall.
   • Height after $n$th bounce: $h_n = 5r^n$.
   • Time to rise (or fall) from $h_n$: $$t_n = \sqrt{\frac{2h_n}{g}} = \sqrt{\frac{2 \times 5r^n}{10}} = \sqrt{r^n} = (0.81)^{\frac{n}{2}} = (0.9)^n$$
   • Since each bounce has an upward and a downward journey, the time for the $n$th bounce is: $$T_n = 2(0.9)^n$$
   • Total time: $$T = t_0 + 2\sum_{n=1}^{\infty} (0.9)^n$$ Using the geometric series formula: $$\sum_{n=1}^{\infty} ar^n = \frac{ar}{1-r}$$ where $a = 1$ and $r = 0.9$, we have: $$\sum_{n=1}^{\infty} (0.9)^n = \frac{0.9}{1-0.9} = \frac{0.9}{0.1} = 9$$ Hence: $$T = 1 + 2 \times 9 = 19 \text{ s}$$

3. Total Distance Traveled:

   • The first drop covers $5$ m.
   • For each bounce, the ball travels upward and downward a distance $5r^n$ each.
   • Total distance: $$D = 5 + 2 \times 5 \sum_{n=1}^{\infty} r^n$$ Using the geometric series formula: $$\sum_{n=1}^{\infty} r^n = \frac{r}{1-r}$$ where $r = 0.81$, we have: $$\sum_{n=1}^{\infty} (0.81)^n = \frac{0.81}{1-0.81} = \frac{0.81}{0.19} \approx 4.263$$ Thus: $$D = 5 + 10 \times 4.263 = 5 + 42.63 = 47.63 \text{ m}$$

4. Average Speed:

   $$\text{Average speed} = \frac{D}{T} = \frac{47.63}{19} \approx 2.50 \text{ m/s}$$

Therefore, the average speed of the ball is approximately 2.50 m/s.