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The problem asks to estimate the instantaneous rate of change in the ball's height at \( t = 3 \) seconds. The function describing the height \( h(t) \) is given as:

\[
h(t) = -5t^2 + 3t + 65
\]

The instantaneous rate of change is the derivative of the height function \( h(t) \) with respect to time \( t \). Let's first find the derivative \( h'(t) \), and then evaluate it at \( t = 3 \).

The derivative of \( h(t) = -5t^2 + 3t + 65 \) is:

\[
h'(t) = \frac{d}{dt} \left(-5t^2 + 3t + 65\right)
\]

Calculating the derivative:

\[
h'(t) = -10t + 3
\]

Now, substituting \( t = 3 \) into the derivative:

\[
h'(3) = -10(3) + 3 = -30 + 3 = -27
\]

Thus, the instantaneous rate of change in the ball's height at \( t = 3 \) seconds is \( -27 \) meters per second.

This means the ball is falling at a rate of 27 meters per second at that moment.

Would you like more details or further clarifications?

### Follow-up Questions:
1. What does the negative sign in the instantaneous rate of change represent in this context?
2. How do we interpret the meaning of the derivative in relation to real-world motion?
3. How would the height of the ball change over time, given the function \( h(t) \)?
4. What are the implications if \( h'(t) = 0 \) at a certain time?
5. How would you compute the average rate of change between two different times?

### Tip:
The derivative gives the instantaneous rate of change, which can be interpreted as the velocity of the object at a particular time in motion problems like this one.

A ball is kicked into the air. Its height is modeled by the function h(t) = -5t^2 + 3t + 65, where h is the height of the object in meters and t is the time in seconds. Estimate the instantaneous rate of change in the ball’s height at 3 s.

Learn how to estimate the instantaneous rate of change for a ball's height using the function h(t) = -5t^2 + 3t + 65. This solution involves using differentiation to find the rate of change at t = 3 seconds. Suitable for students in Grades 11-12.

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