The problem gives us the equation of the height of a rocket as $h = -16t^2 + 75t$, where $t$ is the time in seconds, and $h$ is the height in feet. We are tasked with finding the height of the rocket after 3 seconds.

Step-by-step Solution:

1. Substitute $t = 3$ into the equation: $h = -16(3)^2 + 75(3)$

2. Simplify $3^2$: $h = -16(9) + 75(3)$

3. Multiply: $h = -144 + 225$

4. Combine terms: $h = 81$

Final Answer:

The height of the rocket after 3 seconds is 81 feet.

Would you like further explanation or additional details about the calculation? Here are 5 related questions to deepen understanding:

1. What is the height of the rocket at $t = 0$ seconds (its starting height)?
2. At what time will the rocket reach its maximum height?
3. How would you determine when the rocket hits the ground?
4. What is the physical significance of the coefficients $-16$ and $75$ in the equation?
5. How does the equation relate to the general form of a quadratic equation?

Tip: When solving quadratic equations, always double-check your substitution and operations to avoid errors!