A ball is dropped from a height of 18 m.On each rebound it rises 2/3 of the height from which it last fell. What distance has it travelled at the instant it strikes the ground for the 5th time?

Question

Create a vivid and engaging image illustrating the concept of a ball bouncing. Visualize a bright ball dropping from a significant height, capturing the moments it rebounds higher each time. Show five distinct contact points where the ball interacts with the ground to signify the instances it bounces. Ensure the environment looks clear, perhaps a grass field under a blue sky for contrast. The ball should be visibly compressing and expanding to portray the energy absorption and release during the rebounds. Remember not to include any text in the image.

A ball is dropped from a height of 18 m.On each rebound it rises 2/3 of the height from which it last fell. What distance has it travelled at the instant it strikes the ground for the 5th time?

Answer 1

first strike -- 18 m

2nd strike -- 2(2/3)(18) = (4/3)(18)
3rd strike -- 2(2/3)^2 (18) = 8/9 (18)
4th strike -- 2(2/3)^3 (18) = 16/27 (18)
5th strike -- 2(2/3)^4 (18) = 32/81 (18)

total = 18 + 4 terms of the GP, with a = 72/3 , r = 2/3, n = 4

sum = 18 + (72/3)((1 - 2/3)^4 )/(1 - 2/3)
= appr 75.78 m

Answer 2

Note: The answer differs slightly from the previous one due to rounding.

Answer 3

Well, let's see. The ball is dropping from a height of 18 m, and on each rebound it rises 2/3 of the height from which it last fell. So, after the first rebound, it will rise to a height of (2/3) * 18 = 12 m. Then, it will drop from 12 m, and on the second rebound, it will rise to a height of (2/3) * 12 = 8 m.

So far, we have a total distance of 18 m + 12 m + 8 m = 38 m. Now, on the third rebound, the ball will rise to a height of (2/3) * 8 = 5.33 m. Then, on the fourth rebound, it will rise to a height of (2/3) * 5.33 = 3.56 m.

Adding these distances together, we have: 38 m + 5.33 m + 3.56 m = 46.89 m.

So, at the instant it strikes the ground for the 5th time, the ball will have traveled a distance of approximately 46.89 m. That's quite a journey for a bouncing ball!

Answer 4

To find the distance traveled at the instant the ball strikes the ground for the 5th time, we need to calculate the sum of the distances covered during each rebound.

We can start by calculating the distance covered during the first fall. The ball is dropped from a height of 18 m, so the distance covered during the first fall is 18 m.

Now let's calculate the distances covered during each rebound. The ball rises 2/3 of the height from which it last fell. Since it fell from a height of 18 m, the distance covered during the first rebound would be (2/3) * 18 m = 12 m.

For each subsequent rebound, the ball will also rise 2/3 of the height from which it last fell. So, the distance covered during the second rebound would be (2/3) * 12 m = 8 m, and during the third rebound, it would be (2/3) * 8 m = 5.33 m.

We can continue this pattern for the subsequent rebounds. The distance covered during the fourth rebound would be (2/3) * 5.33 m = 3.56 m, and during the fifth rebound, it would be (2/3) * 3.56 m = 2.37 m.

Now, to find the total distance traveled at the instant the ball strikes the ground for the 5th time, we add up all the distances covered: 18 m + 12 m + 8 m + 5.33 m + 3.56 m + 2.37 m = 49.26 m.

Therefore, the ball has traveled a distance of 49.26 meters at the instant it strikes the ground for the 5th time.

Answer 5

Bounce on each every height from it last fell (Bn x 2/3)

B1 = 18
B2 = 12
B3 = 8
B4 = 16/3
B5 = 32/9

D(total) = H + 2(Btotal)
= 18 + 2(18 + 12 + 8 + 16/3 +32/9) = 75.89 meter.