Need help on a python questions

Question

A certain ball has the property that each time it falls from a height h onto a hard, level surface, it rebounds to a height rh, where r= 4/5. suppose that the ball is dropped from an initial height of 5 meters.

A) Assuming the ball continues to bounce indefinitely, find the total distance it travels.

B) calculate the total time the ball travels.

HINT: use the fact that the ball falls 1/2gt^2 meters in t seconds. use g=9.8 m/s^2

C) If you did not have access to technology, how could you have still determined the sums of the two series you computed for part a and b by hand?

Hint: What type of series were they?

Answer 1

sum = a1/(1-r)= 5/(1-4/5)= 5/1/5 = 25 = sum of all the downward distances

formula for sum of an infinite geometric series

then add the upward distances

25 + 4/1/5 = 25+20 = 45

Answer 2

A) Total Distance Traveled

Let the initial drop height be H=5 m and the rebound ratio \(r=\tfrac45\).

• First drop: the ball falls 5 m.

• Each subsequent bounce: it rises to height 5r^n and then falls the same distance, so contributes 2\cdot5r^n m of travel for n=1,2,3,\dots.

Hence the total distance D is

\[

\begin{aligned}

D &= 5 \;+\; \sum_{n=1}^\infty 2\cdot 5\,r^n \\

&= 5 \;+\; 10\sum_{n=1}^\infty r^n

= 5 \;+\; 10\!\left(\frac{r}{1-r}\right)

= 5 \;+\; 10\!\left(\frac{\tfrac45}{1-\tfrac45}\right)

= 5 + 10\cdot 4

= 45\text{ m}.

\end{aligned}

\]

B) Total Time Traveled

A fall (or rise) from height h takes t=\sqrt{\frac{2h}{g}} with g=9.8 m/s².

1. Initial drop time:

t_0 = \sqrt{\frac{2\cdot5}{9.8}}.

2. Each bounce cycle (rise + fall from height 5r^n):

2\;\sqrt{\frac{2\,(5r^n)}{9.8}} =2\,\sqrt{\frac{2\cdot5}{9.8}}\;\sqrt{r^n}.

So the total time T is

\begin{aligned} T &= \sqrt{\frac{2\cdot5}{9.8}} \;\Bigl[\,1 \;+\; 2\sum_{n=1}^\infty r^{n/2}\Bigr] \\ &= \sqrt{\frac{10}{9.8}} \;\Bigl[\,1 \;+\; 2\!\bigl(\frac{\sqrt r}{1-\sqrt r}\bigr)\Bigr]. \end{aligned}

Numerically,

T \approx 1.01015\times\bigl[\,1 + 2\cdot\frac{\sqrt{4/5}}{1-\sqrt{4/5}}\bigr] \approx 18.1\text{ s}.

C) Summing “By Hand” Without Technology

Both parts lead to infinite geometric series:

• Distance series: \sum_{n=1}^\infty r^n

• Time series: \sum_{n=1}^\infty \bigl(\sqrt r\,\bigr)^n

You use the closed‑form formula for an infinite geometric series,

\sum_{n=0}^\infty ar^n = \frac{a}{1-r}\quad(|r|<1),

(or its shifted version starting at n=1), to compute each sum exactly.

Answer 3

Let's simplify this by taking the distance in two parts.

The first distance is the initial drop: 5 m

The second distance is the sum of the bounces. The first bounce is 4/5 up and 4/5 down: 8 m

a₁ = 8

r = 4 / 5

S = a₁ / (1 - r)

S = 8 / (1 - (4/5))

S = 1.6

Total distance is 5 + 40 meters

Answer 4

I am not sure whether this is a python programming question, or a calculus question.

I will first answer part c, i.e. the calculus portion.

Let me summarize the facts I assume you have been given.

Fact 1:

1 + r + r² + r³ + ... is a geometric series.

If |r| < 1 then the series converges to the quantity 1/(1-r)

Fact 2:

If a series is absolutely convergent

(i.e., it is converges even if we take absolute values of the individual terms)

then its sum remains the same after permutation of the terms.

Fact 3:

After a ball rebounces, it falls back to the ground taking the same about of time it took to bounce up.

If the height is x then the time is √(2x/g) (using your teacher's hint).

Now let's get to your ball.

During the first fall it

travels distance h

taking time √(2h/g).

During the first rise it

travels distance hr

taking time √(2hr/g).

During the second fall, it takes the same distance and time as during the previous rise: It

travels distance hr

taking time √(2hr/g).

During the second rise it

travels distance hr²

taking time √(2hr²/g).

During the third fall, it takes the same distance and time as during the previous rise: It

travels distance hr²

taking time √(2hr²/g).

And so on.

So the total distance travelled is

h + hr + hr + hr² + hr² + hr³ + hr³ + ... =

h + 2hr + 2hr² + 2hr³ + ... =

h + 2hr(1 + r + r² + r³ + ...) =

h + 2hr/(1-r)

The total time is

√(2h/g) + √(2hr/g) + √(2hr/g) + √(2hr²/g) + √(2hr²/g) + √(2hr³/g) + √(2hr³/g) + ...

√(2h/g) + 2√(2hr/g) + 2√(2hr²/g) + 2√(2hr³/g) + ...

√(2h/g) + 2√(2hr/g) (1 + √r + √r² + ...)

As this infinite series is absolutely convergent, we can rearrange the terms like so

1 + √r + √r² + ... =

(1 + √r² + √r⁴ + ...) + (√r + √r³ + ...) =

(1 + r + r² + ...) + √r(1 + r + r² + ...) =

(1 + r + r² + ...)(1 + √r) =

(1 + √r)/(1-r)

Thus the total time is

√(2h/g) + 2√(2hr/g)(1 + √r)/(1-r) =

√(2h/g)(1 + 2√r(1 + √r)/(1-r) )

If you implement it, and print the resulting times and distances,

you will see that they converge to the above calculated quantities.