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Step-by-Step Detailed Solution
Step 1: Understand the Problem
The ball is dropped from a height of 5 m. Each time it hits the floor, it bounces back to 81/100 (i.e., 0.81) of the previous height. We are to find its average speed over the entire motion, assuming the process continues indefinitely. We take the acceleration due to gravity to be g = 10 m/s2.
Step 2: Calculate the Total Distance Traveled
First drop: The ball travels 5 m downward on its first drop.
Subsequent bounces: After the first bounce, the ball rises to a maximum height of $0.81 \times 5 = 4.05\,\text{m}$ and then falls the same distance. Each bounce thereafter is $0.81$ times the height of the previous bounce.
General form of total distance:
The total distance $D_\text{total}$ is given by:
$$
D_\text{total} = 5 \;+\; 2 \times 5 \times 0.81 \;+\; 2 \times 5 \times (0.81)^2 \;+\; 2 \times 5 \times (0.81)^3 \;+\; \dots
$$
This forms an infinite geometric series for all the rises and falls after the initial drop.
Sum of the series:
The infinite sum of distances after the first drop is:
$$
2 \times 5 \sum_{n=1}^{\infty} (0.81)^n
\;=\; 10 \sum_{n=1}^{\infty} (0.81)^n
\;=\; 10 \left( \frac{0.81}{1 - 0.81} \right)
\;=\; 10 \left( \frac{0.81}{0.19} \right).
$$
Numerically,
$$
10 \times \frac{0.81}{0.19} \approx 42.63.
$$
Total distance traveled:
Therefore,
$$
D_\text{total} = 5 + 42.63 \;\approx\; 47.63\,\text{m}.
$$
Step 3: Calculate the Total Time Taken
Time for the first drop:
The time to fall 5 m under gravity (with $g = 10\,\text{m/s}^2$) is
$$
t_0 = \sqrt{\frac{2 \times 5}{10}} = \sqrt{1} = 1\,\text{s}.
$$
Velocity right before hitting the ground the first time:
$$
v = \sqrt{2gh}
= \sqrt{2 \times 10 \times 5}
= \sqrt{100}
= 10\,\text{m/s}.
$$
Time for each bounce up and down:
After each bounce, the speed is multiplied by $0.9$ (since $ \sqrt{0.81} = 0.9$). For the nth bounce, the rebound speed is $10 \times (0.9)^n$. The time for going up (or down) at the nth bounce is
$$
\frac{10 \times (0.9)^n}{g} = \frac{10 \times (0.9)^n}{10} = (0.9)^n.
$$
The total time for the nth bounce (up and down) is $2 \times (0.9)^n$.
Total bounce time:
Summing this for $n = 1$ to $\infty$:
$$
\sum_{n=1}^{\infty} 2 \times (0.9)^n
= 2 \sum_{n=1}^{\infty} (0.9)^n
= 2 \left( \frac{0.9}{1 - 0.9} \right)
= 2 \times 9
= 18\,\text{s}.
$$
Total time of motion:
Hence, the total time
$$
T = t_0 + 18 = 1 + 18 = 19\,\text{s}.
$$
Step 4: Compute the Average Speed
The average speed $v_\text{avg}$ is the total distance divided by the total time:
$$
v_\text{avg} = \frac{D_\text{total}}{T}
= \frac{47.63}{19}
\approx 2.50\,\text{m/s}.
$$
Step 5: Final Answer
The average speed of the ball is approximately 2.50 m/s.
