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The height \( h \) of the water balloon is given by the equation:

\[
h = 20t - 5t^2
\]

To find the values of \( t \) and \( h \) when the balloon is at its release point, we need to consider the initial conditions right when the balloon is launched. At this release point, \( t = 0 \) seconds, because it’s the starting moment of the motion.

### Step 1: Substitute \( t = 0 \) into the equation
\[
h = 20(0) - 5(0)^2 = 0
\]

### Conclusion
The values at the release point are:
\[
t = 0, \quad h = 0
\]

Would you like further details or have any questions?

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Here are five related questions to explore this concept further:

1. How can we determine the maximum height of the water balloon and at what time it occurs?
2. What would be the height of the balloon at \( t = 2 \) seconds?
3. How can we find the time when the balloon hits the ground again?
4. What does the coefficient \(-5\) in the equation indicate about the balloon's trajectory?
5. How would this equation change if air resistance was taken into account?

**Tip:** For quadratic equations modeling motion, the maximum point often represents the peak height in projectile motion.

What are the values of t and h when the water balloon is at its release point based on the quadratic equation h = 20t - 5t^2?

Learn how to determine the time and height of a water balloon at its release point using the quadratic equation h = 20t - 5t^2. This problem involves key concepts in quadratic functions and projectile motion. Ideal for students in Grades 7-9.

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