Math Problem Statement

A rocket is fired upward from some initial distance above the ground. Its height (in feet), h, above the ground t seconds after it is fired is given by h(t) = -16t^2 + 96t + 640. What is the rocket's maximum height? How long does it take for the rocket to reach its maximum height? After it is fired, the rocket reaches the ground at t = ? seconds.

Solution

The given height equation is h(t)=16t2+96t+640h(t) = -16t^2 + 96t + 640, which represents the height of the rocket in feet as a function of time tt seconds.

1. Maximum Height of the Rocket:

To find the maximum height, we need to determine the vertex of the quadratic equation. The time tt at which the maximum height occurs is given by the formula for the vertex of a parabola:

t=b2at = \frac{-b}{2a}

Here, a=16a = -16, b=96b = 96, and c=640c = 640 (though cc is not needed for this step).

t=962(16)=9632=3 secondst = \frac{-96}{2(-16)} = \frac{-96}{-32} = 3 \text{ seconds}

Now, substitute t=3t = 3 back into the original equation to find the maximum height:

h(3)=16(3)2+96(3)+640h(3) = -16(3)^2 + 96(3) + 640 h(3)=16(9)+288+640=144+288+640=784 feeth(3) = -16(9) + 288 + 640 = -144 + 288 + 640 = 784 \text{ feet}

So, the rocket's maximum height is 784 feet.

2. Time to Reach Maximum Height:

We have already calculated that the rocket reaches its maximum height at 3 seconds.

3. Time to Reach the Ground:

To find when the rocket reaches the ground, we set h(t)=0h(t) = 0 and solve for tt:

16t2+96t+640=0-16t^2 + 96t + 640 = 0

We can solve this quadratic equation using the quadratic formula:

t=b±b24ac2at = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Substituting the values of a=16a = -16, b=96b = 96, and c=640c = 640:

t=96±9624(16)(640)2(16)t = \frac{-96 \pm \sqrt{96^2 - 4(-16)(640)}}{2(-16)} t=96±9216+4096032t = \frac{-96 \pm \sqrt{9216 + 40960}}{-32} t=96±5017632t = \frac{-96 \pm \sqrt{50176}}{-32} t=96±22432t = \frac{-96 \pm 224}{-32}

Now, calculate the two possible values for tt:

t1=96+22432=12832=4 seconds (not possible, discard this value)t_1 = \frac{-96 + 224}{-32} = \frac{128}{-32} = -4 \text{ seconds (not possible, discard this value)} t2=9622432=32032=10 secondst_2 = \frac{-96 - 224}{-32} = \frac{-320}{-32} = 10 \text{ seconds}

So, the rocket reaches the ground after 10 seconds.


Summary of Answers:

  • The rocket's maximum height is 784 feet.
  • It takes 3 seconds to reach the maximum height.
  • The rocket reaches the ground after 10 seconds.

Would you like further details or have any questions?


Follow-up Questions:

  1. What is the initial height of the rocket when t=0t = 0?
  2. How fast is the rocket moving when it hits the ground?
  3. What is the velocity of the rocket at its maximum height?
  4. Can you calculate the time when the rocket is halfway to its maximum height?
  5. How would the equation change if the rocket was launched with a different initial velocity?

Tip:

To determine the velocity of an object from its height function, take the derivative of the height function with respect to time.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Projectile Motion

Formulas

Quadratic equation: ax^2 + bx + c = 0
Vertex formula for a parabola: t = -b / 2a
Quadratic formula: t = [-b ± sqrt(b^2 - 4ac)] / 2a

Theorems

Quadratic formula
Properties of Parabolas

Suitable Grade Level

Grades 9-11