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+80t+576. 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

To analyze the rocket's motion, we are given the height equation:

h(t)=16t2+80t+576h(t) = -16t^2 + 80t + 576

Step 1: Find the time to reach maximum height

The time to reach the maximum height occurs at the vertex of the parabola, which is given by:

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

Here, a=16a = -16, b=80b = 80, and c=576c = 576. Substituting these values:

t=802(16)=8032=2.5seconds.t = -\frac{80}{2(-16)} = \frac{80}{32} = 2.5 \, \text{seconds}.

So, the rocket reaches its maximum height at t=2.5t = 2.5 seconds.


Step 2: Find the maximum height

To find the maximum height, substitute t=2.5t = 2.5 into the height equation:

h(2.5)=16(2.5)2+80(2.5)+576h(2.5) = -16(2.5)^2 + 80(2.5) + 576

Calculate step by step:

h(2.5)=16(6.25)+80(2.5)+576h(2.5) = -16(6.25) + 80(2.5) + 576

h(2.5)=100+200+576=676feet.h(2.5) = -100 + 200 + 576 = 676 \, \text{feet}.

The maximum height is 676feet676 \, \text{feet}.


Step 3: Find when the rocket reaches the ground

The rocket reaches the ground when h(t)=0h(t) = 0. Solve:

16t2+80t+576=0-16t^2 + 80t + 576 = 0

Divide the entire equation by 16-16:

t25t36=0t^2 - 5t - 36 = 0

Factorize or use the quadratic formula. The quadratic formula is:

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

Here, a=1a = 1, b=5b = -5, c=36c = -36. Substituting:

t=(5)±(5)24(1)(36)2(1)t = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-36)}}{2(1)}

t=5±25+1442t = \frac{5 \pm \sqrt{25 + 144}}{2}

t=5±1692t = \frac{5 \pm \sqrt{169}}{2}

t=5±132t = \frac{5 \pm 13}{2}

t=182=9ort=82=4.t = \frac{18}{2} = 9 \quad \text{or} \quad t = \frac{-8}{2} = -4.

Since time cannot be negative, t=9secondst = 9 \, \text{seconds}.


Final Results:

  1. Maximum height: 676feet676 \, \text{feet}.
  2. Time to reach maximum height: 2.5seconds2.5 \, \text{seconds}.
  3. Time to hit the ground: 9seconds9 \, \text{seconds}.

Do you want a detailed explanation of any step or graphing? Here are some related questions:

  1. How does changing the initial height affect the results?
  2. What happens to the maximum height if the initial velocity is doubled?
  3. Can we graph h(t)h(t) to visualize the motion of the rocket?
  4. How would the results change if the equation included air resistance?
  5. How can we calculate the velocity of the rocket at a specific time?

Tip: Always check the discriminant (b24acb^2 - 4ac) before solving quadratic equations to confirm real solutions.

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Math Problem Analysis

Mathematical Concepts

Quadratic Equations
Vertex of a Parabola
Factoring
Roots of Quadratic Functions

Formulas

h(t) = at^2 + bt + c
t = -b / 2a
Quadratic Formula: t = (-b ± √(b^2 - 4ac)) / 2a

Theorems

Vertex Formula for Quadratic Functions
Quadratic Formula

Suitable Grade Level

Grades 9-12