Parabolic Functions Homework Worksheet - Word Problems on Projectile Motion

Problem Analysis and Solution

The provided image contains three problems involving quadratic functions, which model the motion of objects under gravity. We will solve each problem step by step.

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#### Problem 1: Rocket Launch

The height \( h \) (in meters) of a rocket above the ground as a function of time \( t \) (in seconds) is given by:
\[
h = -5t^2 + 20t
\]

##### a. What is the rocket's maximum height?

To find the maximum height, we need to determine the vertex of the parabola represented by the quadratic equation \( h = -5t^2 + 20t \). The vertex form of a quadratic equation \( h = at^2 + bt + c \) has its maximum (or minimum) value at:
\[
t = -\frac{b}{2a}
\]
Here, \( a = -5 \) and \( b = 20 \). Plugging in these values:
\[
t = -\frac{20}{2(-5)} = -\frac{20}{-10} = 2
\]
So, the rocket reaches its maximum height at \( t = 2 \) seconds.

Next, we substitute \( t = 2 \) into the height equation to find the maximum height:
\[
h = -5(2)^2 + 20(2) = -5(4) + 40 = -20 + 40 = 20
\]
Thus, the maximum height is:
\[
\boxed{20 \text{ meters}}
\]

##### b. When does the rocket reach the maximum height?

From the calculation above, the rocket reaches its maximum height at:
\[
\boxed{2 \text{ seconds}}
\]

##### c. When does the rocket hit the ground?

The rocket hits the ground when its height \( h \) is 0. We solve the equation:
\[
-5t^2 + 20t = 0
\]
Factor out \( t \):
\[
t(-5t + 20) = 0
\]
This gives two solutions:
\[
t = 0 \quad \text{or} \quad -5t + 20 = 0
\]
Solving the second equation:
\[
-5t + 20 = 0 \implies -5t = -20 \implies t = 4
\]
The solution \( t = 0 \) corresponds to the initial launch, so the rocket hits the ground at:
\[
\boxed{4 \text{ seconds}}
\]

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#### Problem 2: Water Balloon

The height \( h \) (in meters) of a water balloon thrown upward is given by:
\[
h = -4.9t^2 + 27t + 2
\]

##### a. How high is the balloon after 1 second?

Substitute \( t = 1 \) into the height equation:
\[
h = -4.9(1)^2 + 27(1) + 2 = -4.9(1) + 27 + 2 = -4.9 + 27 + 2 = 24.1
\]
Thus, the height of the balloon after 1 second is:
\[
\boxed{24.1 \text{ meters}}
\]

##### b. Find the roots of the parabola.

The roots of the parabola are the values of \( t \) for which \( h = 0 \). Solve the equation:
\[
-4.9t^2 + 27t + 2 = 0
\]
This is a quadratic equation of the form \( at^2 + bt + c = 0 \), where \( a = -4.9 \), \( b = 27 \), and \( c = 2 \). Use the quadratic formula:
\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Substitute the values:
\[
t = \frac{-27 \pm \sqrt{27^2 - 4(-4.9)(2)}}{2(-4.9)}
\]
\[
t = \frac{-27 \pm \sqrt{729 + 39.2}}{-9.8}
\]
\[
t = \frac{-27 \pm \sqrt{768.2}}{-9.8}
\]
\[
t = \frac{-27 \pm 27.716}{-9.8}
\]
This gives two solutions:
\[
t = \frac{-27 + 27.716}{-9.8} = \frac{0.716}{-9.8} \approx -0.073 \quad (\text{not physically meaningful})
\]
\[
t = \frac{-27 - 27.716}{-9.8} = \frac{-54.716}{-9.8} \approx 5.583
\]
The physically meaningful root is:
\[
t \approx 5.583 \text{ seconds}
\]
Thus, the roots of the parabola are:
\[
\boxed{t \approx 5.583 \text{ seconds}}
\]

##### c. When will the balloon hit the ground?

The balloon hits the ground when \( h = 0 \), which we have already solved. The relevant root is:
\[
\boxed{5.583 \text{ seconds}}
\]

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#### Problem 3: Football

The height \( h \) (in meters) of a football thrown is given by:
\[
h = -4.9t^2 + 24.5t + 1
\]

##### a. How high is the ball after 1.4 seconds?

Substitute \( t = 1.4 \) into the height equation:
\[
h = -4.9(1.4)^2 + 24.5(1.4) + 1
\]
First, calculate \( 1.4^2 \):
\[
1.4^2 = 1.96
\]
Now substitute:
\[
h = -4.9(1.96) + 24.5(1.4) + 1
\]
\[
h = -9.604 + 34.3 + 1
\]
\[
h = 25.696
\]
Thus, the height of the ball after 1.4 seconds is:
\[
\boxed{25.696 \text{ meters}}
\]

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Final Answers:

1. Rocket Launch:
- Maximum height: \(\boxed{20 \text{ meters}}\)
- Time to reach maximum height: \(\boxed{2 \text{ seconds}}\)
- Time to hit the ground: \(\boxed{4 \text{ seconds}}\)

2. Water Balloon:
- Height after 1 second: \(\boxed{24.1 \text{ meters}}\)
- Roots of the parabola: \(\boxed{t \approx 5.583 \text{ seconds}}\)
- Time to hit the ground: \(\boxed{5.583 \text{ seconds}}\)

3. Football:
- Height after 1.4 seconds: \(\boxed{25.696 \text{ meters}}\)