After striking the floor, a rubber ball rebounds to 4/5thof the height from which it has fallen. Find the total distance that it travels before coming to rest if it has been gently dropped from a height of 120 metres.a)540 mb)960 mc)1080 md)1020 me)1120 mCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

The first drop is 120 metres. After this the ball will rise by 96 metres and fall by 96 metres. This process will continue in the form of infinite GP with common ratio 0.8 and first term 96. The required answer is given by the formula:

a/(1-r)

Now,

[{120/(1/5)}+{96/(1/5)}]

= 1080 m.

Answer

How's the answer 1080 ?

Answer

Answer

1080

Answer

► So, starting from a height of 120m, the object will rebound to 4/5th of its original height after striking the floor each and every time.
► This means after the first drop the ball will rebound and will fall with that rebound metres, and so on.

► We get a series from these observations

Total distance travelled = 120 + 96 + 96 + 76.8 + 76.8 + 61.4 + 61.4 + ….

⇒ 120 + 2*(96 + 76.8 + 61.4 …)

⇒ 120 + 2*(96 + 96*(4/5) + 96(4/5)² + …)

► Now, inside those brackets, it is a Geometric Progression or a GP with the first term a = 96 and the common ratio r = 4/5 = 0.8

► As it is an infinite GP, the sum of all it's terms is equal to a / (1-r)

So, the sum of distances covered = 120 + 2*96 / (1 - 0.8)
⇒ 120 + 960 = 1080m.

Answer

To find the total distance that the rubber ball travels before coming to rest, we need to consider the distance it travels on each bounce.

Let's assume the initial height from which the ball is dropped is H = 120 meters.

The ball rebounds to 4/5th of the height from which it has fallen, which means it reaches a height of (4/5) * H = (4/5) * 120 = 96 meters on each bounce.

We can calculate the distance traveled by the ball using the formula for the sum of an infinite geometric series:

S = a / (1 - r)

Where:
- S is the total distance traveled
- a is the first term in the series (the initial height)
- r is the common ratio (the ratio of the height on each bounce to the previous height)

Let's calculate the total distance traveled by the ball.

First Term:
The ball is initially dropped from a height of 120 meters, so the first term (a) is 120 meters.

Common Ratio:
The ball rebounds to 4/5th of the height from which it has fallen, so the common ratio (r) is 4/5.

Total Distance Traveled:
S = 120 / (1 - 4/5)
S = 120 / (1/5)
S = 120 * 5
S = 600 meters

However, this calculation only considers the distance traveled on the ascents (when the ball is bouncing up). To find the total distance traveled, we need to consider the descents as well.

Since the ball rebounds to 4/5th of the height from which it has fallen, it also descends to 4/5th of the previous height on each descent.

Therefore, the total distance traveled is twice the distance calculated above:

Total Distance Traveled = 2 * S
Total Distance Traveled = 2 * 600
Total Distance Traveled = 1200 meters

But be careful! The question asks for the total distance traveled before the ball comes to rest. The ball comes to rest when it reaches a height of 0.

Since the ball starts at a height of 120 meters and reaches a height of 96 meters on each bounce, there will be a point where the ball does not rebound to a positive height.

To find the number of bounces before the ball comes to rest, we can set up the following inequality:

(4/5) ^ n * 120 < />

Simplifying the inequality, we get:

(4/5) ^ n < />

Since any positive number raised to any power will always be positive, the inequality is not possible. Therefore, the ball will never reach a height of 0 and will continue bouncing indefinitely.

Therefore, the total distance traveled by the ball before coming to rest is infinite.

None of the answer options provided (a, b, c, d, e) are correct.

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