Bouncing Ball

Algebra Level 2

After being dropped, a ball always bounces back to $\frac{2}{5}$ of the height of its previous bounce. After the first bounce, it reaches a height of 200 inches. How high will it reach after the tenth bounce?

5.24E-2 0.232 0.678 0.897

2 solutions

Hung Woei Neoh
Jun 22, 2016

The height reached after the $n$ -th bounce is actually a geometric progression with $a=200$ and $r=\dfrac{2}{5} = 0.4$

Therefore, the height after the $10$ th bounce:

$T_{10} = 200(0.4)^{10-1} = 200(0.4)^9 = \boxed{0.0524}$

Nice solution, honestly better than mine :)

Hana Wehbi - 4 years, 11 months ago

Our solutions are actually the same...

Hung Woei Neoh - 4 years, 11 months ago

But you are right, it is a geometric progression and it was good to mention it.

Hana Wehbi - 4 years, 11 months ago

Hana Wehbi
Jun 22, 2016

If after each bounce, the ball reaches $\frac{2}{5}$ of the previous height. Thus, after the tenth bounce, it will reach

( $\frac{2}{5}$ ) $^{9}\times 200$ = $\color{#3D99F6}{\boxed{0.0524}}$

Bouncing Ball

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