Question

question 5 of 10
a ball is dropped from a height of 12 feet and returns to a height that is one-half of the height from which it fell. the ball continues to bounce half the height of the previous bounce each time. how far will the ball have traveled vertically when it hits the ground for the fourth time? it might be helpful to sketch the path of the ball.

a. 24 feet
b. 33 feet
c. 3 feet
d. 12 feet

Explanation:

Step1: Analyze the first drop

The ball is dropped from 12 feet, so the first drop distance is 12 feet.

Step2: Analyze the first bounce up and down

After the first drop, it bounces up to \( \frac{12}{2} = 6 \) feet and then falls back down 6 feet. So this contributes \( 6 + 6 = 12 \) feet.

Step3: Analyze the second bounce up and down

Next, it bounces up to \( \frac{6}{2}=3 \) feet and then falls back down 3 feet. This contributes \( 3 + 3 = 6 \) feet.

Step4: Analyze the third bounce up (before the fourth hit)

Before the fourth hit, it bounces up to \( \frac{3}{2} = 1.5 \) feet, but we only care about the distance until the fourth hit. Wait, no, let's re - track the hits:

• First hit: after falling 12 feet.
• Second hit: after bouncing up 6 feet and falling 6 feet (total so far: \( 12+6 + 6=24 \) feet, second hit).
• Third hit: after bouncing up 3 feet and falling 3 feet (total so far: \( 24+3 + 3 = 30 \) feet, third hit). Wait, no, maybe a better way:

The sequence of distances:

• Drop 1: 12 feet (first hit)
• Bounce up 1: 6 feet, drop 1: 6 feet (second hit, total \( 12 + 6+6=24 \))
• Bounce up 2: 3 feet, drop 2: 3 feet (third hit, total \( 24+3 + 3=30 \))
• Bounce up 3: 1.5 feet, drop 3: 1.5 feet (fourth hit, total \( 30+1.5 + 1.5=33 \))

Wait, no, maybe my initial step - by - step was wrong. Let's list the events:

1. First drop: 12 feet (hits ground first time)
2. Bounce up: 6 feet, then drop 6 feet (hits ground second time, total distance: \( 12+6 + 6 = 24 \))
3. Bounce up: 3 feet, then drop 3 feet (hits ground third time, total distance: \( 24+3 + 3=30 \))
4. Bounce up: 1.5 feet, then drop 1.5 feet (hits ground fourth time, total distance: \( 30 + 1.5+1.5=33 \))

Wait, but maybe the problem is considering the path as:

First drop: 12

Then, each bounce - up and bounce - down (except the last bounce - up before the fourth hit). Wait, the question is "when it hits the ground for the fourth time". Let's count the number of falls:

• Fall 1: 12 (hit 1)
• Fall 2: 6 (after bouncing up 6) (hit 2)
• Fall 3: 3 (after bouncing up 3) (hit 3)
• Fall 4: 1.5 (after bouncing up 1.5) (hit 4)

And the number of bounces - up:

• Bounce up 1: 6 (before fall 2)
• Bounce up 2: 3 (before fall 3)
• Bounce up 3: 1.5 (before fall 4)

So total distance is \( 12+(6 + 6)+(3 + 3)+(1.5+1.5) \)? No, no, when it hits the ground for the fourth time, the number of falls is 4, and the number of bounces - up is 3.

Wait, let's do it correctly:

The formula for the total distance \( D \) when hitting the ground \( n \) times:

For a ball dropped from height \( h \), with each bounce height \( r\times h \) (where \( r=\frac{1}{2} \) here):

• When \( n = 1 \), \( D=h \)
• When \( n = 2 \), \( D=h+2\times h\times r \)
• When \( n = 3 \), \( D=h + 2\times h\times r+2\times h\times r^{2}\)
• When \( n = 4 \), \( D=h+2\times h\times r + 2\times h\times r^{2}+2\times h\times r^{3}\)

Here, \( h = 12 \), \( r=\frac{1}{2} \)

So \( D=12+2\times12\times\frac{1}{2}+2\times12\times(\frac{1}{2})^{2}+2\times12\times(\frac{1}{2})^{3}\)

Calculate each term:

• First term: 12
• Second term: \( 2\times12\times\frac{1}{2}=12 \)
• Third term: \( 2\times12\times\frac{1}{4}=6 \)
• Fourth term: \( 2\times12\times\frac{1}{8}=3 \)

Sum them up: \( 12 + 12+6 + 3=33 \) feet.

Answer:

B. 33 feet