A model rocket is fired from the side of a hill and its flight is given by the quadratic equation shown here:

Question

A model rocket is fired from the side of a hill and its flight is given by the quadratic equation shown here:

ℎ(𝑡𝑡) = −4.9𝑡𝑡2 + 50𝑡𝑡 + 25
a) What is the initial height of the rocket? (1 mark)
b) What is the maximum height of the rocket? (1 mark)
c) How long after launch does the rocket reach the maximum height? (1 mark)
d) How long does it take for the rocket to reach the ground? (3 marks)
e) State the domain and range for this situation.

Answer 1

a) The initial height of the rocket is given by the constant term in the equation, which is 25.

b) The maximum height of the rocket can be found by determining the vertex of the quadratic equation. The formula for finding the x-coordinate of the vertex of a quadratic equation in the form 𝑓(𝑥) = 𝑎𝑥^2 + 𝑏𝑥 + 𝑐 is given by 𝑥 = -𝑏/(2𝑎). In this case, 𝑎 = -4.9 and 𝑏 = 50.
So, 𝑡 = -50/(2*(-4.9)) ≈ 5.102.
To find the maximum height, substitute this value into the equation: ℎ(𝑡) = −4.9(5.102)^2 + 50(5.102) + 25 ≈ 125.375.
Therefore, the maximum height of the rocket is approximately 125.375.

c) The time at which the rocket reaches the maximum height is given by the x-coordinate of the vertex, which we found to be 𝑡 = 5.102. Therefore, the rocket reaches its maximum height approximately 5.102 seconds after launch.

d) To find the time it takes for the rocket to reach the ground, we need to find the time at which the height ℎ(𝑡) equals zero. To do this, set ℎ(𝑡) = 0 in the quadratic equation and solve for 𝑡.
-4.9𝑡^2 + 50𝑡 + 25 = 0
Using the quadratic formula, we get 𝑡 = (-50 ± sqrt(50^2 - 4*(-4.9)*25))/(2*(-4.9)).
Calculating this, we find that 𝑡 ≈ -1.02 and 𝑡 ≈ 5.13.
Since the time cannot be negative in this context, we discard the 𝑡 ≈ -1.02 solution.
Therefore, it takes approximately 5.13 seconds for the rocket to reach the ground.

e) The domain of this situation is the set of all real numbers, as time (𝑡) can be any real number.
The range is the set of all real numbers greater than or equal to the initial height of the rocket, which is 25.

Answer 2

. Explain how the coordinates of the vertex help you find the maximum or minimum value of a quadratic

and where that maximum or minimum occurs.

Answer 3

The coordinates of the vertex of a quadratic equation in the form 𝑦 = 𝑎𝑥^2 + 𝑏𝑥 + 𝑐, where 𝑎 ≠ 0, help us find the maximum or minimum value of the quadratic and where that maximum or minimum occurs.

The x-coordinate of the vertex, denoted as 𝑥 = -𝑏/(2𝑎), gives us the value of 𝑥 at which the quadratic reaches its maximum or minimum. If 𝑎 is positive, the vertex represents the minimum point, and if 𝑎 is negative, the vertex represents the maximum point.

The y-coordinate of the vertex, denoted as 𝑦 = 𝑓(𝑥), gives us the value of 𝑦 at the maximum or minimum point. In other words, it gives us the maximum or minimum value of the quadratic.

By calculating the coordinates of the vertex, we can determine the maximum or minimum value and the corresponding 𝑥-value of the quadratic. This information is useful in various real-life situations, such as determining the maximum height of a projectile or optimizing the production of a product to maximize profits.

Answer 4

a) The initial height of the rocket can be obtained by evaluating ℎ(0) in the quadratic equation.

ℎ(0) = -4.9(0)^2 + 50(0) + 25
= 0 + 0 + 25
= 25

Therefore, the initial height of the rocket is 25.

b) The maximum height of the rocket can be determined by finding the vertex of the quadratic equation.
The vertex of the equation ℎ(𝑡𝑡) = -4.9𝑡𝑡^2 + 50𝑡𝑡 + 25 can be found using the formula ℎ(𝑡𝑡) = -𝑏𝑏/2𝑎𝑎, where 𝑎 = -4.9 and 𝑏 = 50:
ℎ(𝑡𝑡) = -50/(2(-4.9))
ℎ(𝑡𝑡) = -50/(-9.8)
ℎ(𝑡𝑡) = 5.1

Therefore, the maximum height of the rocket is 5.1.

c) The time it takes for the rocket to reach the maximum height can be determined by finding the time when the rocket reaches its vertex using the formula -𝑏𝑏/2𝑎𝑎.
𝑡𝑡 = -50/(2(-4.9))
𝑡𝑡 ≈ 5.10 seconds

Therefore, the rocket reaches its maximum height approximately 5.10 seconds after launch.

d) To determine the time it takes for the rocket to reach the ground, we need to find the time when ℎ(𝑡𝑡) = 0.
Setting ℎ(𝑡𝑡) = 0 in the quadratic equation and solving for 𝑡𝑡:
-4.9(𝑡𝑡)^2 + 50(𝑡𝑡) + 25 = 0

Using the quadratic formula, 𝑡𝑡 = (-𝑏𝑏 ± √(𝑏𝑏^2 - 4𝑎𝑎𝑐𝑐)) / 2𝑎𝑎:
𝑡𝑡 = (-50 ± √(50^2 - 4(-4.9)(25))) / (2(-4.9))
𝑡𝑡 ≈ (-50 ± √(2500 + 4900)) / (-9.8)
𝑡𝑡 ≈ (-50 ± √(7400)) / (-9.8)
𝑡𝑡 ≈ (-50 ± 86.02) / (-9.8)
𝑡𝑡 ≈ (-50 - 86.02) / (-9.8) or 𝑡𝑡 ≈ (-50 + 86.02) / (-9.8)
𝑡𝑡 ≈ 13.07 seconds or 𝑡𝑡 ≈ -1.57 seconds

The rocket does not take -1.57 seconds to reach the ground, so the answer is 13.07 seconds.

Therefore, it takes approximately 13.07 seconds for the rocket to reach the ground.

e) The domain for this situation is the set of all real numbers since time (𝑡𝑡) can be any real number.
The range for this situation is the set of all real numbers greater than or equal to the initial height (25), since 𝑡𝑡 can be any real number but the height cannot be negative.

Answer 5

To find the answers to these questions, we'll need to use some concepts of quadratic equations. Let's break down each question and explain how to solve it.

a) What is the initial height of the rocket?
The initial height of the rocket can be found by evaluating the equation ℎ(𝑡) at 𝑡 = 0. Substituting 𝑡 = 0 into the equation ℎ(𝑡) = −4.9𝑡^2 + 50𝑡 + 25, we get ℎ(0) = −4.9(0)^2 + 50(0) + 25. Simplifying this, we find the initial height of the rocket.

b) What is the maximum height of the rocket?
The maximum height of the rocket corresponds to the vertex of the quadratic equation. The x-coordinate of the vertex can be found using the formula 𝑡 = −𝑏/2𝑎, where 𝑎, 𝑏, and 𝑐 are coefficients of the quadratic equation ℎ(𝑡) = 𝑎𝑡^2 + 𝑏𝑡 + 𝑐. In this case, 𝑎 = −4.9 and 𝑏 = 50. Substituting these values into the formula, we can determine 𝑡. Then, substituting this value of 𝑡 back into the equation ℎ(𝑡), we can find the maximum height.

c) How long after launch does the rocket reach the maximum height?
The time it takes to reach the maximum height is 𝑡, which we found in the previous step.

d) How long does it take for the rocket to reach the ground?
To find how long it takes for the rocket to reach the ground, we need to solve the quadratic equation ℎ(𝑡) = 0. This can be done by factoring, completing the square, or using the quadratic formula. Once we have the solutions, we determine the positive value, as negative time values are not relevant in this context.

e) State the domain and range for this situation.
The domain refers to the set of all possible input values (𝑡-values) for the equation ℎ(𝑡). In this case, since the rocket is being launched, we assume that 𝑡 starts at 0 and can continue indefinitely.

The range refers to the set of all possible output values (ℎ-values) for the equation ℎ(𝑡). Since the rocket is fired vertically and can theoretically reach any positive height, the range of this situation is all real numbers greater than or equal to the initial height.