Class 11 Mathematicas

Question

A bouncing tennis ball rebounds each time to a height one half the height of the previous bounce. It is dropped from a height of 10m. Find:

Answer 1

To find the total distance traveled by the bouncing tennis ball, we need to consider the sum of all the distances it travels during each bounce, including the initial drop.

Given:

- The ball is initially dropped from a height of 10 meters.

- Each time it bounces, it rebounds to a height one-half of the previous bounce.

Let's break down the distances for each bounce:

1. The initial drop: 10 meters (h₁ = 10 m)

Now, for each subsequent bounce:

2. First bounce:

- The ball rebounds to a height of half the initial drop height: h₁/2 = 10/2 = 5 meters.

- It also travels the same distance on the way up and down, so the total distance traveled during the first bounce is 2 * 5 = 10 meters.

3. Second bounce:

- The ball rebounds to a height of half of the previous bounce height: h₂ = (h₁/2)/2 = (10/2)/2 = 2.5 meters.

- Again, it travels the same distance on the way up and down, so the total distance traveled during the second bounce is 2 * 2.5 = 5 meters.

4. Third bounce:

- The ball rebounds to a height of half of the previous bounce height: h₃ = (h₂/2)/2 = (2.5/2)/2 = 1.25 meters.

- It travels the same distance on the way up and down, so the total distance traveled during the third bounce is 2 * 1.25 = 2.5 meters.

You can see that this pattern continues for each subsequent bounce, with the height of the bounce halving each time. The total distance traveled can be calculated as the sum of the distances for each bounce:

Total distance = Initial drop distance + Sum of distances during bounces

Total distance = 10 + (10 + 5 + 2.5 + ...) meters

Now, we can calculate the sum of the infinite geometric series:

S = a / (1 - r)

Where:

- a is the first term (initial drop distance) = 10 meters

- r is the common ratio (1/2)

S = 10 / (1 - 1/2)

S = 10 / (1/2)

S = 20 meters

So, the total distance traveled by the bouncing tennis ball is 20 meters.

Answer 2

1st bounce 5 m

2nd 2.5 m

3rd 1.25 m

4th 0.625 m

5th 0.3125 m

6th 0.15625 m

etc

ad infinitum

xth 10/2^(x-1) meters

getting ever closer to no bounce but never reaching 0 m as minimum bounce

it's a geometric sequence

x=a1(r^n-1) with a1=10 m, 4 + 1/2, n=x= number of bounces

x =10/(1/2)^(x-1)=10/2^(x-1)

S = a1/(1 -r) = 10/1/2 = 20 m for an infinite series

for total distance traveled at the End of time

or = the limit which is approached but never reached, like an asymptote

Sx =a1(1-r^x)/(1-r) for a finite series

such as x =6 bounces,

S6 =10(1-(1/2)^6)/(1-1/2) = 20/63/64= 5/63/16=315/16 m

=10+5+2.5+1.25+.625 +.3125

=10+5 + 2 1/2+1 1/4 + 5/8+5/16=19 11/16=315/16