Merchant Taylors’ School Specimen Paper 1 Q36

For our final question in the Merchant Taylors’ specimen paper, we delve into a classic problem about bouncing balls. If you need the full paper, check Merchant Taylors’ website. We might also have a link on our question page or our YouTube page.

The Problem Statement

A ball is dropped and bounces up to a height that is 75% of the height from which it was dropped. It then bounces again to a height that is 75% of the previous height and continues in this manner.

The Big Question

How many bounces does it make before it bounces to a height that is less than 25% of the original height?

This is a significant reduction, and understanding the process requires careful analysis.

Visualizing the Problem

Let’s start by visualizing the motion of the ball:

1. Initially, the ball is at its original height.

   ![Ball_Height_Initial](

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2. It drops to the floor and bounces back up to 75% of its original height.

   ![First_Bounce_Height](

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3. On subsequent bounces, it continues to reach 75% of the height from which it last bounced.

   ![Subsequent_Bounces](

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Calculating the Heights

To determine how quickly the ball’s height decreases, let’s track the height after each bounce:

1. First Bounce

   • The ball reaches 75% of its original height.

   • Best understood as:[\text{New Height} = 0.75 \times \text{Original Height}]

     • Example: If the original height is 100 meters, after the first bounce, it reaches:[0.75 \times 100 = 75 \text{ meters}]

2. Second Bounce

   • Same principle applies,

   • Now, it reaches 75% of the previously attained height:[\text{New Height} = 0.75 \times 75%]

     • Example: Continuing from the first bounce height, after the second bounce:[0.75 \times 75 \text{ meters} = 56.25 \text{ meters}]

Proceeding Further

3. Third Bounce

   • Following the pattern:[\text{New Height} = 0.75 \times 56.25 (\text{previous height from second bounce})]

     • This calculation gives approximately:[0.75 \times 56.25 = 42.19 \text{ meters}]

Let’s continue this pattern a bit:

[\begin{align*}\text{Fourth Bounce} & : 0.75 \times 42.19 \approx 31.64 \text{ meters}\\text{Fifth Bounce} & : 0.75 \times 31.64 \approx 23.73 \text{ meters}\end{align*}]

Notice how the heights are progressively decreasing.

Setting Up the Mathematics

To find how many bounces it makes before the ball’s height is less than 25% of the original height, we need:

[0.75^{n} < 0.25]

Where ( n ) is the number of bounces. To solve for ( n ), we can use logarithms:

[n \log(0.75) < \log(0.25)]

Solving for ( n ):

[n > \frac{\log(0.25)}{\log(0.75)}]

Using a calculator, you find:

[\log(0.25) \approx -0.602 \quad \text{and} \quad \log(0.75) \approx -0.125]

Thus,

[n > \frac{-0.602}{-0.125} \approx 4.816]

Conclusion

The closest whole number greater than 4.816 is 5. So, the ball must bounce 5 times to reach a height less than 25% of its original height.

Key Points

• After 5 bounces, the ball height goes under 25% of the original height.

• Understanding the decay of height involves calculating ( 0.75^{n} ) and comparing it with the threshold.

Here’s a summarized progression:

1. Original Height: 100%

2. 1st Bounce: 75%

3. 2nd Bounce: 56.25%

4. 3rd Bounce: 42.19%

5. 4th Bounce: 31.64%

6. 5th Bounce: 23.73% < 25%

Thus, 5 bounces are necessary.

“A ball’s resilience fades as quickly as it bounces, teaching us about persistence and diminishing returns with each effort.”

Let us know how you tackled the problem and share your thoughts in the comments below!

Stay tuned for more insights on problem-solving techniques and exam strategies.

[ \text{Happy Studying!} ]

[ \text{Example Calculation visual} ]

![Example_Calculation](

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