a ball is dropped from a height of 48m and rebounds two-third of the distance it falls. If it continues to fall and rebound in this way,how far will it travel before coming to rest?

Make a little diagram of what actually happens.

After the first bounce, the ball goes up and down 2/3 of it previous bounce, but the first event can be considered only half a bounce.

so I am going to consider this to be a geometric series, pretending the first bounce is complete. Just have to remember to subtract 48 from the final answer.

so sum = 96 + 96(2/3) + 96(2/3)^2 + ...
a = 96, r = 2/3

sum(all the terms) = a/(1-r)
= 96/(1- 2/3)
= 96/(1/3)
= 288
But this includes the first 48 m, which did not happen.

total distance = 288-48 = 240 m

or

total distance
= 48 + 2(48)(2/3) + 2(48)(2/3)^2 + ...
= 48 + 96(2/3)[ 1 + 2/3 + (2/3)^2 + ...]
= 48 + 64 [1/(1-2/3)]
= 48 + 64[1/3]
= 240

To find how far the ball will travel before coming to rest, we can calculate the sum of the distances it travels while falling and rebounding.

Given:
Initial height (h) = 48 m
Rebound distance ratio = 2/3

Step 1: Calculate the distance the ball travels on the first fall.
Distance of the first fall = Initial height (h) = 48 m

Step 2: Calculate the distance of the first rebound.
Distance of the first rebound = Rebound distance ratio * distance of the first fall
Distance of the first rebound = (2/3) * 48 m = 32 m

Step 3: Calculate the distance the ball travels on the second fall, using the starting point of the first rebound.
Distance of the second fall = Distance of the first rebound
Distance of the second fall = 32 m

Step 4: Calculate the distance of the second rebound.
Distance of the second rebound = Rebound distance ratio * distance of the second fall
Distance of the second rebound = (2/3) * 32 m = 21.33 m (rounding to two decimal places)

Step 5: Calculate the distance the ball travels on the third fall, using the starting point of the second rebound.
Distance of the third fall = Distance of the second rebound
Distance of the third fall = 21.33 m

Step 6: Calculate the distance of the third rebound.
Distance of the third rebound = Rebound distance ratio * distance of the third fall
Distance of the third rebound = (2/3) * 21.33 m = 14.22 m (rounding to two decimal places)

Continue this process until the distance of the rebound becomes negligible (approaching zero).

Now, we need to find the sum of all the distances traveled by the ball.

Sum of distances = Distance of first fall + Distance of first rebound + Distance of second fall + Distance of second rebound + Distance of third fall + Distance of third rebound + ...

Sum of distances = 48 m + 32 m + 32 m + 21.33 m + 21.33 m + 14.22 m + 14.22 m + ...

To find the sum, we can use the formula for the sum of an infinite geometric series:

Sum = a / (1 - r)

Where:
a = Distance of first fall = 48 m
r = Rebound distance ratio = 2/3

Sum = 48 m / (1 - 2/3)

Simplifying,
Sum = 48 m / (1/3)
Sum = 48 m * (3/1)
Sum = 48 m * 3
Sum = 144 m

Therefore, the ball will travel a total distance of 144 meters before coming to rest.

To find the total distance the ball will travel before coming to rest, we need to sum up the distances it covers in each fall-rebound cycle.

The ball is dropped from a height of 48m, so in the first fall it covers 48m. It then rebounds two-thirds of this distance (2/3 * 48 = 32m) and falls again. In the second fall, it covers 32m, and rebounds two-thirds of this distance (2/3 * 32 = 21.33m). This process continues indefinitely until the ball comes to rest.

To determine the total distance the ball will travel, we can use the geometric series formula:

Sum = first term / (1 - common ratio)

In this case, the first term is 48m and the common ratio is 2/3.

Sum = 48 / (1 - 2/3)
= 48 / (1/3)
= 48 * (3/1)
= 144m

Therefore, the ball will travel a total distance of 144 meters before coming to rest.