A rubber ball is dropped from a height of 9m and each time it strikes the ground, pounces to a height 2/3 of that from which it alls. Find the total distance travelled by the all before it comes to rest.

Answer 1

To find the total distance travelled by the ball before it comes to rest, we need to sum up the distances it travels during each bounce.

The ball is dropped from a height of 9m, so it travels a distance of 9m during the first fall.

After the first bounce, the ball reaches a height of (2/3) * 9m = 6m.
Then, it falls again and reaches a height of 6m during the second fall.

After the second bounce, the ball reaches a height of (2/3) * 6m = 4m.
Then, it falls and reaches a height of 4m during the third fall.

After the third bounce, the ball reaches a height of (2/3) * 4m = 8/3 m.
Then, it falls and reaches a height of 8/3 m during the fourth fall.

For each subsequent bounce, the height reached decreases by a factor of 2/3.

Let's calculate the distances travelled during the bounces and the falls:

1st fall: 9m
1st bounce: 9m
2nd fall: 6m
2nd bounce: 6m
3rd fall: 4m
3rd bounce: 4m
4th fall: 8/3 m
4th bounce: 8/3 m
5th fall: (2/3) * (8/3) m = 16/9 m

The total distance travelled by the ball is:
9m + 9m + 6m + 6m + 4m + 4m + 8/3 m + 8/3 m + 16/9 m

We can simplify this expression:
9m + 9m + 6m + 6m + 4m + 4m + 8/3 m + 8/3 m + 16/9 m = 36m + 16/3 m + 16/9 m

Now, we need a common denominator to add these fractions:
36m + 16/3 m + 16/9 m = (108/9)m + (48/9)m + 16/9 m = (108 + 48 + 16)/9 m = 172/9 m

The total distance travelled by the ball before it comes to rest is 172/9 m.