5. A rubber ball dropped on a hard surface takes a sequence of bounces, each one 3/5 as high as the preceding one. If this ball is dropped from a height of 15 feet, what is the total distance it has traveled after it hits the surface the fifth time?
The answer is 54 and 21/125 ft, but why?
A rubber ball dropped on a hard surface takes a sequence of bounces, each one 3/5 as high as the preceding one. If this ball is dropped from a height of 15 feet, what is the total distance it has traveled after it hits the surface the fifth time?
Dwn.Up.Dwn.Up.Dwn..Up...Dwn...Up...Dwn
15..9...9.5.4.5.4.3.24.3.24.1.944.1.944
Ssumming yields 54.168 = 54 21/125 feet.
why are u answering a question with another question ??? ;-;
It's a sort og geometric series with limiting sum. Please I need a perfect formula to calculate it instead of the normal addition
To find the total distance traveled by the ball after it hits the surface for the fifth time, we need to consider the distance covered during each bounce and add them up.
Let's analyze the bounces:
1st bounce: The ball is dropped from a height of 15 feet, so it travels a distance of 15 feet.
2nd bounce: The ball bounces to 3/5 of the height of the previous bounce. Therefore, it reaches a height of (3/5) * 15 feet and falls back, covering a distance of (3/5) * 15 feet.
3rd bounce: The ball bounces to 3/5 of the height of the previous bounce, which is (3/5) * (3/5) * 15 feet.
4th bounce: The ball bounces to 3/5 of the height of the previous bounce, which is (3/5) * (3/5) * (3/5) * 15 feet.
5th bounce: The ball bounces to 3/5 of the height of the previous bounce, which is (3/5) * (3/5) * (3/5) * (3/5) * 15 feet.
To find the total distance traveled, we need to sum up the distances covered during each bounce:
Total distance = 15 + (3/5) * 15 + (3/5) * (3/5) * 15 + (3/5) * (3/5) * (3/5) * 15 + (3/5) * (3/5) * (3/5) * (3/5) * 15
To simplify the calculation, let's look at the pattern:
(3/5)^0 * 15 + (3/5)^1 * 15 + (3/5)^2 * 15 + (3/5)^3 * 15 + (3/5)^4 * 15
Using the formula for the sum of a geometric series, we can simplify this expression:
Total distance = 15 * [(1 - (3/5)^5) / (1 - (3/5))]
Total distance = 15 * [(1 - 0.07776) / (0.2)]
Total distance = 15 * [0.92224 / 0.2]
Total distance = 15 * 4.6112
Total distance = 69.1672 feet
Therefore, the total distance traveled by the ball after it hits the surface for the fifth time is approximately 69.1672 feet.
It seems there might be an error in the given answer of 54 and 21/125 feet. Double-check the calculations to confirm the correct answer.
each time a bouncy ball bounce, it bounces to half the height from which it had just fallen. if jack drops a ball from a building and it rises to a height of 24 metres, how high will it bounce on the fifth bounce?
Guys, here is the answer.
Ball is dropped from a height of 15 feet and each bounce it goes to 3/5 as high as the preceding one.
Now, total distance traveled in:
1st drop:15 (till it hits the ground)
2nd drop:(15*3/5)=9*2=18 [Multiply by 2 as distance to go up (15*3/5) and to come down (15*3/5),so multiply by 2]
3rd drop: (9*3/5)=5.4*2 =10.8
4th drop: (5.4*3/5)=3.24*2 =6.48
5th drop: (3.24*3/5)=1.94*2= 3.888 (now ball will hit the ground after 5th fall)
total distance: distance in 1st drop+ 2nd drop+3rd drop+4th drop+5th drop.
= 15+18+10.8+6.48+3.888=54.168
Hope this helps....