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.