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Let's work through the problems visible in the image.

### Problem (g) and (h):
**(g)**: \( x^2 + y = 3 \), where \( x \) is the input and \( y \) is the output.  
You can solve for \( y \) in terms of \( x \) to make this function clearer:
\[
y = 3 - x^2
\]
So the output \( y \) depends on subtracting the square of the input \( x \) from 3.

**(h)**: \( 1 + y^2 = x^2 \), where \( x \) is the input and \( y \) is the output.  
Rearranging this equation to solve for \( y \):
\[
y^2 = x^2 - 1
\]
\[
y = \pm \sqrt{x^2 - 1}
\]
This gives two possible outputs, \( +\sqrt{x^2 - 1} \) or \( -\sqrt{x^2 - 1} \), depending on the value of \( x \).

### Problem 2:
The height of the ball at a horizontal distance \( x \) feet from the thrower is given by the function:
\[
h(x) = -\frac{32}{400}x^2 + x + 5
\]

#### (a) Find the height of the ball when it is 5 feet from the thrower.
We are asked to find \( h(5) \). Substituting \( x = 5 \) into the equation for \( h(x) \):
\[
h(5) = -\frac{32}{400}(5)^2 + 5 + 5
\]
Simplifying:
\[
h(5) = -\frac{32}{400}(25) + 5 + 5
\]
\[
h(5) = -2 + 5 + 5 = 8
\]
So, the height of the ball when it is 5 feet from the thrower is **8 feet**.

#### (b) At what height is the ball thrown from?
The height at which the ball is thrown corresponds to the value of \( h(x) \) when \( x = 0 \) (right at the point the ball leaves the thrower's hand):
\[
h(0) = -\frac{32}{400}(0)^2 + 0 + 5 = 5
\]
So, the ball is thrown from a height of **5 feet**.

#### (c) If the other player is 5 feet tall, how far away should the other player stand to catch the ball?
We need to find \( x \) when the height of the ball equals 5 feet, i.e., solve \( h(x) = 5 \):
\[
5 = -\frac{32}{400}x^2 + x + 5
\]
Simplifying:
\[
0 = -\frac{32}{400}x^2 + x
\]
Multiply through by 400 to eliminate the fraction:
\[
0 = -32x^2 + 400x
\]
Factor out \( x \):
\[
x(400 - 32x) = 0
\]
So, \( x = 0 \) (at the thrower) or \( x = \frac{400}{32} = 12.5 \).

Thus, the other player should stand **12.5 feet** away to catch the ball at a height of 5 feet.

---

Would you like a detailed explanation of any part of the solution? Here are 5 related questions for deeper exploration:
1. How would the height equation change if the ball was thrown at a higher initial position?
2. How does the term \( -\frac{32}{400}x^2 \) affect the shape of the trajectory?
3. What are the physical interpretations of the solutions in part (c)?
4. How would the answer to (a) change if the thrower was standing on a hill 3 feet higher?
5. How can we use the quadratic formula to solve problems like part (c)?

**Tip**: For quadratic motion problems, always check if factoring or using the quadratic formula is possible to simplify solving for distances or times.

Solve the following problems:

1. (g) x^2 + y = 3, where x is the input and y is the output.
   (h) 1 + y^2 = x^2, where x is the input and y is the output.

2. A baseball player throws a ball across the field. The height of the ball at a horizontal distance x feet from the thrower is given by the function:
   h(x) = -32/400x^2 + x + 5
   (a) Find the height of the ball when it is 5 feet from the thrower.
   (b) At what height is the ball thrown from?
   (c) If the other player is 5 feet tall, how far away should the other player stand to catch the ball?

This problem set includes solving quadratic equations and analyzing parabolic motion. Topics covered include algebraic functions and calculating a ball’s trajectory. Suitable for students in grades 9-12.

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