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The given quadratic equation for the height \( h \) of the rocket is:

\[
h = -16t^2 + 160t
\]

where:
- \( h \) is the height in meters,
- \( t \) is the time in seconds.

We are asked to find the height after \( t = 1.25 \) seconds.

Substitute \( t = 1.25 \) into the equation:

\[
h = -16(1.25)^2 + 160(1.25)
\]

Let's calculate it step by step:

\[
h = -16(1.5625) + 160(1.25)
\]
\[
h = -25 + 200
\]
\[
h = 175 \text{ meters}
\]

So, the rocket's height after 1.25 seconds is **175 meters**.

Do you want further details or have any questions?

Here are five related questions you could explore:
1. How would the height change if the rocket were launched with a different initial velocity?
2. At what time does the rocket reach its maximum height?
3. How long does it take for the rocket to return to the ground?
4. What would happen to the rocket's height if the acceleration due to gravity changed?
5. How does the quadratic term \( -16t^2 \) affect the rocket's trajectory?

**Tip:** In quadratic motion problems, the coefficient of \( t^2 \) represents the effect of gravity or acceleration on the motion.

The height, h, in metres, of a rocket t seconds after it is launched is approximately modelled by the quadratic relation h = -16t^2 + 160t. After 1.25 seconds, to the nearest metre, how high is the rocket?

Learn how to calculate the height of a rocket after 1.25 seconds using the quadratic equation h = -16t^2 + 160t. This example demonstrates key concepts in quadratic equations and projectile motion, ideal for students in Grades 9-11.

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