We are given the formula for the hang time $t$, which is expressed as:

$t = \frac{\sqrt{d}}{2}$

where $d$ is the vertical distance (in feet) and $t$ is the hang time (in seconds). We are asked to find the vertical distance $d$ when the player's hang time is $t = 1.23$ seconds.

Steps to solve for $d$:

1. Start with the equation: $t = \frac{\sqrt{d}}{2}$ Substitute $t = 1.23$: $1.23 = \frac{\sqrt{d}}{2}$

2. Multiply both sides by 2 to isolate $\sqrt{d}$: $2 \times 1.23 = \sqrt{d}$ $\sqrt{d} = 2.46$

3. Square both sides to solve for $d$: $d = (2.46)^2$ $d = 6.0516$

4. Round $d$ to the nearest tenth: $d \approx 6.1 \, \text{feet}$

Final Answer:

The vertical distance $d$ of the player's jump is approximately 6.1 feet.

Would you like any further details or clarifications?

Here are 5 related questions:

1. How would the formula change if the hang time were measured in meters instead of feet?
2. What is the maximum vertical distance if the hang time was 2 seconds?
3. If another player's hang time is 1.5 seconds, what is their jump height?
4. How does the hang time change if the vertical distance doubles?
5. What is the physical significance of the factor $\frac{1}{2}$ in the hang time equation?

Tip: When solving equations involving square roots, always remember to check whether you need to consider both positive and negative roots, depending on the context of the problem.