A ball is dropped from a platform $20 \mathrm{~m}$ high. On each successive bounce, the ball looses $25 \%$ of its previous height. a) What total distance will the ball have travelled after $16^{\text {th }}$ bounce (just before it hits the ground for the $17^{\text {th }}$ time)? b) What total distance will the ball have travelled when it stops? a) Total distance after $16^{\text {th }}$ bounce is: Number $\mathrm{m}$. (provide your answer rounded to 2 decimal places if applicable) b) Total distance by the time the ball stops is: Number $\mathrm{m}$. (provide your answer rounded to 2 decimal places if applicable)

Step 1 ：Let's find the total distance traveled by the ball after the 16th bounce and when it stops. We know that the ball loses 25% of its height on each bounce. We can use the geometric series formula to calculate the total distance.

Step 2 ：For the 16th bounce, we can calculate the sum of the first 16 terms of the geometric series with the initial height of 20m and a common ratio of 0.75 (since it loses 25% of its height).

Step 3 ：The formula for the sum of a geometric series is: \(S_n = \frac{a_1(1 - r^n)}{1 - r}\) where \(S_n\) is the sum of the first n terms, \(a_1\) is the initial height, \(r\) is the common ratio, and \(n\) is the number of terms.

Step 4 ：For the total distance when the ball stops, we need to find the sum of the infinite geometric series. The formula for the sum of an infinite geometric series is: \(S_\infty = \frac{a_1}{1 - r}\)

Step 5 ：We can plug in the values and calculate the total distance after the 16th bounce and when the ball stops: initial_height = 20, common_ratio = 0.75, n_terms = 16

Step 6 ：Calculating the sum of the first 16 terms: \(S_{16} = \frac{20(1 - 0.75^{16})}{1 - 0.75} = 79.19819233939052\)

Step 7 ：Calculating the sum of the infinite geometric series: \(S_\infty = \frac{20}{1 - 0.75} = 80.0\)

Step 8 ：Final Answer: a) The total distance after the $16^\text{th}$ bounce is: \(\boxed{79.20 \mathrm{~m}}\) (rounded to 2 decimal places).

Step 9 ：b) The total distance by the time the ball stops is: \(\boxed{80.00 \mathrm{~m}}\) (rounded to 2 decimal places).