Math Problem Statement

An object is released into the air at an height of 12ft and upward velocity of 30ft per second. Write a function to model height h, in feet, of the object t seconds after it was released.

Solution

To model the height hh of the object tt seconds after it was released, you can use the following quadratic function derived from the principles of projectile motion:

h(t)=16t2+v0t+h0h(t) = -16t^2 + v_0 t + h_0

Where:

  • h(t)h(t) is the height in feet at time tt.
  • v0v_0 is the initial upward velocity (30 ft/s).
  • h0h_0 is the initial height (12 ft).
  • The term 16t2-16t^2 accounts for the effect of gravity (approximately 32ft/s232 \, \text{ft/s}^2, hence divided by 2).

Substituting the values:

h(t)=16t2+30t+12h(t) = -16t^2 + 30t + 12

This function gives the height of the object at any time tt after it has been released.

Would you like further details or have any questions? Here are some related questions for you to consider:

  1. What is the maximum height the object reaches?
  2. How long does it take for the object to hit the ground?
  3. What is the height of the object after 1 second?
  4. How does changing the initial velocity affect the height function?
  5. What would happen if the object was thrown downward instead of upward?

Tip: When modeling projectile motion, always remember to account for both the initial height and velocity in your function.

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Math Problem Analysis

Mathematical Concepts

Projectile Motion
Quadratic Functions

Formulas

h(t) = -16t^2 + v_0 t + h_0

Theorems

Principles of projectile motion

Suitable Grade Level

Grades 8-10