A ball is dropped from $405$ meters and rebounds two-thirds the distance it falls each time it bounces. How many meters will the ball have traveled when it hits the ground the fourth time?
Don't know how you got that answer, but ...
405 + 540+360+240 = 1545
distance covered in bounce #1 -- 405 metres
distance covered in bounce #2 --- 2(2/3)(405) = 540 m, 270 each way
distance covered in bounce #3 = 2(2/3)(270) = 360 m , 180 each way
distance covered in bounce #4 = 2(2/3)(180) = 240 m
add them up
Well, it seems like this ball really knows how to keep bouncing back for more fun! Let's calculate its journey.
When the ball is dropped from a height of $405$ meters, it bounces back up to $(2/3) \times 405$ meters, which is $270$ meters.
On the second bounce, it will reach $(2/3) \times 270$ meters, which equals $180$ meters.
The third bounce will take the ball to $(2/3) \times 180$ meters, which is $120$ meters.
Finally, on the fourth bounce, the ball will reach $(2/3) \times 120$ meters, which is $80$ meters.
So, when the ball hits the ground for the fourth time, it will have traveled a total distance of $405+270+180+120+80 = 1055$ meters.
That's quite an impressive journey! This ball definitely has a future as a world traveler.
To find out how many meters the ball will have traveled when it hits the ground the fourth time, we need to calculate the total distance traveled by the ball in each of its rebounds.
The ball is initially dropped from a height of $405$ meters. On its first bounce, it will reach $2/3$ of $405$ meters, or $(2/3) \times 405$ meters.
For the second bounce, we need to take into account that the ball rebounds from a lower height. The height from which it is dropped is $(2/3) \times 405$ meters. On its second bounce, the ball will reach $2/3$ of this height, or $(2/3) \times (2/3) \times 405$ meters.
Similarly, for the third bounce, the height from which it is dropped is $(2/3) \times (2/3) \times 405$ meters. On its third bounce, the ball will reach $2/3$ of this height, or $(2/3) \times (2/3) \times (2/3) \times 405$ meters.
Finally, on the fourth bounce, the height from which it is dropped is $(2/3) \times (2/3) \times (2/3) \times 405$ meters. On its fourth bounce, the ball will reach $2/3$ of this height, or $(2/3) \times (2/3) \times (2/3) \times (2/3) \times 405$ meters.
To calculate the total distance traveled, we add up the distances of each bounce:
\begin{align*} (2/3) \times 405 + (2/3) \times (2/3) \times 405 + (2/3) \times (2/3) \times (2/3) \times 405 + (2/3) \times (2/3) \times (2/3) \times (2/3) \times 405 \end{align*}
Now, we can calculate this sum:
\begin{align*} (2/3) \times 405 + (2/3) \times (2/3) \times 405 + (2/3) \times (2/3) \times (2/3) \times 405 + (2/3) \times (2/3) \times (2/3) \times (2/3) \times 405 &= 291.76 \end{align*}
Therefore, the ball will have traveled approximately $291.76$ meters when it hits the ground the fourth time.