Think logically not matheatically [part-23]

Classical Mechanics Level 2

A bouncy ball is dropped from a tower of height 100m. then it bounces back $\frac{1}{4}$ th of its initial path then the maximum displacement travelled by it after it comes to rest in the ground is _ _ _ _ _

25m

\frac{300}{4}

\frac{400}{3}

m 100m

3 solutions

Sudoku Subbu
Feb 17, 2015

If i asked the distance then $\frac{400}{3}$ is the correct answer but i asked the displacement so the answer is 100m don't create a dispute this is one of my logical problems visit my set for more logical problems

I. Failed haaaaahaaahaa....I am one of that foolish guys who took it for distance.....so a nice questions........ :-)

ashutosh mahapatra - 6 years, 3 months ago

Me too hahaha

Abyoso Hapsoro - 6 years, 1 month ago

Just for kicks, how would you solve for the distance traveled?

I'm getting a distance of $\displaystyle \frac{500}{3} m$ , would you mind telling me how you got your answer of $\displaystyle \frac{400}{3} m$ ?

Here's my process: 100 meters is the initial drop height, so that's added once.

Then the ball comes back up by 25m, but every time it comes back up, it must go back down and travel the same distance down as it did up. So in this case, it goes back down 25m. Therefore, I multiply the sum of the infinite geometric series by 2, with common ratio $\displaystyle \frac{1}{4}$ .

My result ends up being $\displaystyle 100+ 2 \times \sum_{n=0}^\infty (25m \times ( \frac{1}{4})^n)$ , which, using the infinite sum formula for a geometric series, for $\displaystyle -1<r<1$

$\displaystyle S_\infty= \frac{a_0}{1-r}$

which returns a final answer of $\displaystyle \frac{500}{3} m$ .

Chris M. - 6 years ago

yeah by the formula $\frac{a}{1-r}$ where a=10 and r= $\frac{1}{4}$ $distance=\frac{100}{1-\frac{1}{4}}$ $=\frac{100}{\frac{3}{4}}$ $=100\times\frac{4}{3}$ $=\frac{400}{3}$

sudoku subbu - 5 years, 11 months ago

I'm afraid I don't quite understand how your solution compensates for the fact that the ball will bounce back up, increasing the distance traveled, resulting in the ball's distance traveled being something like this:

$\displaystyle 100m (down) + 25m (up) + 25m (down) + \frac{25}{4}m (up) + \frac{25}{4}m (down)+...$

Or, by doubling the distances where the ball travels both up and down: $\displaystyle 100m (down) + 2(25m (down) + \frac{25}{4}m (down)+...)$

The 100m is not within the doubled portion since the ball begins at a position 100 meters above the ground, and will only be able to bounce back up a fraction of that amount (1/4, 1/16, 1/64, etc.) as its energy is dissipated.

The total distance would then be the initial 100 meters plus twice the sum of a geometric series, with $a=25m$ and $\displaystyle r=\frac {1}{4}$

$\displaystyle distance = 100m + 2(\frac{25m}{1-\frac{1}{4}}) = 100m+\frac{50}{\frac{3}{4}}m=\frac{500}{3} m$

Chris M. - 5 years, 10 months ago

Chris M.
May 31, 2015

If we were prompted for the total distance traveled, we would solve it using the sum of an infinite geometric series, however, we were asked for the ball's displacement. This would simply be the difference between its final position and initial position, which would be 100m (assuming down is defined as the positive direction and the ground is considered 0).

Nelson Mandela
Feb 17, 2015

so finally when the ball comes to rest, the total displacement is height = 100m.

Think logically not matheatically [part-23]

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