Math Problem Statement

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?

Solution

The given quadratic equation for the height hh of the rocket is:

h=16t2+160th = -16t^2 + 160t

where:

  • hh is the height in meters,
  • tt is the time in seconds.

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

Substitute t=1.25t = 1.25 into the equation:

h=16(1.25)2+160(1.25)h = -16(1.25)^2 + 160(1.25)

Let's calculate it step by step:

h=16(1.5625)+160(1.25)h = -16(1.5625) + 160(1.25) h=25+200h = -25 + 200 h=175 metersh = 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 16t2-16t^2 affect the rocket's trajectory?

Tip: In quadratic motion problems, the coefficient of t2t^2 represents the effect of gravity or acceleration on the motion.

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Math Problem Analysis

Mathematical Concepts

Quadratic Equations
Projectile Motion

Formulas

h = -16t^2 + 160t (height of rocket)

Theorems

Kinematic Equations for Uniformly Accelerated Motion

Suitable Grade Level

Grades 9-11