The Problem:
Create a quadratic function to model the height of an object thrown into the air with the following conditions
The maximum height is between 90 and 100 m
The object is starts on the ground
The object takes at least 10 s before returning to the ground
1 mark for providing data
1 mark for completing an equation with a "maximum" between 90 to 100.
1 mark for showing work for the calculation of the maximum
1 mark for giving an equation that has the object starting on the ground
1 mark for showing work for the calculation of the initial height
1 mark for giving an equation that has a second zero of at least 10
1 mark for showing work for the second zero
To create a quadratic function, we need to find the vertex (the maximum point) and two other points on the path. Let's choose the vertex to be (5, 95) since this is halfway between the starting and ending point and within our maximum height range. The other two points will be the initial point (0, 0) and the point when the object returns to the ground, let's say (10, 0).
A quadratic function has the form y = a(x-h)^2 + k, where (h, k) is the vertex. Thus, the equation will be y = a(x - 5)^2 + 95.
Now we need to find the value of 'a'. We can use one of the other two points for this, let's use (0, 0).
0 = a(0 - 5)^2 + 95
-95 = 25a
a = -3.8
So our equation becomes y = -3.8(x - 5)^2 + 95. Since this quadratic goes downward, we know the maximum height occurs at the vertex, which we chose to be (5, 95). Now we just plug in the initial point and the point when the object returns to the ground to make sure our equation works for both.
Initial height:
y = -3.8(0 - 5)^2 + 95 = 0
Second zero:
y = -3.8(10 - 5)^2 + 95 = 0
Thus, our quadratic function that models the height of the object is y = -3.8(x - 5)^2 + 95.
To create a quadratic function that models the given conditions, we will start by considering the properties of a quadratic function in vertex form, which is given by:
f(x) = a(x - h)^2 + k
where (h, k) represents the vertex of the parabola.
1. Finding the maximum height:
To have a maximum height between 90 and 100 m, we can let the vertex be at (h, k) = (0, 95). This ensures that the vertex is within the desired range.
Therefore, our equation becomes:
f(x) = a(x - 0)^2 + 95
= ax^2 + 95
2. Ensuring the object starts on the ground:
To make sure the object starts on the ground, we need the initial height to be zero. So, our equation becomes:
f(x) = ax^2 + 95 + 0
= ax^2 + 95
3. Calculating the value of 'a':
To calculate the value of 'a', we can use the condition that the object takes at least 10 seconds before returning to the ground. This means that the x-coordinate of the second zero of the quadratic function should be at least 10.
When the object hits the ground, f(x) = 0.
Plugging in x = 10 and setting the equation to zero:
0 = a(10)^2 + 95
0 = 100a + 95
-95 = 100a
a = -95/100
a = -0.95
Therefore, our final quadratic function is:
f(x) = -0.95x^2 + 95
If you have any further questions, please, let me know.
To create a quadratic function that models the height of the object thrown into the air, we need to consider the given conditions.
1. Maximum height between 90 and 100 m:
Let's assume the maximum height is represented by the coordinate (x, y) where y represents the maximum height. Since the maximum height occurs at the vertex of a quadratic function, we need to determine the x-coordinate of the vertex to satisfy the condition.
The x-coordinate of the vertex is given by the formula:
x = -b / (2a)
To find the x-coordinate, we need to determine the coefficients a, b, and c of the quadratic function equation of the form y = ax^2 + bx + c.
Since the maximum height occurs at the vertex, the axis of symmetry passes through the vertex. Therefore, the time it takes for the object to reach the maximum height is equal to half the time it takes for the object to return to the ground.
Considering the object starts on the ground and takes at least 10 seconds before returning to the ground:
The duration from the start to the maximum height is half the total time, so it takes at least 5 seconds to reach the maximum height. Therefore, we will set t = 5.
Now, let's assume the height at the maximum point is 95m (midway between 90 and 100 m).
Substituting these values into the general equation of motion for vertical displacement h(t):
h(t) = -16t^2 + vt + h0
Where h(t) represents the height at time t, v represents the initial upwards velocity, and h0 represents the initial height.
Using the known values from before:
95 = -16(5)^2 + 5v + h0
This equation represents the first condition of the height of the maximum point.
2. Object starting on the ground:
Since the object starts on the ground, the initial height, h0, is 0. Substituting this value into the equation above, we can solve for v:
95 = -16(5)^2 + 5v + 0
95 = -400 + 5v
5v = 495
v = 99
So, the initial upwards velocity, v, is 99 m/s.
Now, we can finalize the equation of motion:
h(t) = -16t^2 + 99t
This equation satisfies the first two conditions.
3. Second zero of at least 10:
To calculate the second zero, we set h(t) equal to 0 and solve for t:
-16t^2 + 99t = 0
t(-16t + 99) = 0
This equation has two solutions: t = 0 (initial time when the object is thrown) and -16t + 99 = 0.
Solving -16t + 99 = 0:
-16t = -99
t = 99/16
Since time cannot be negative, we disregard the t = 0 solution. The t = 99/16 solution gives us the second zero.
Therefore, the quadratic function that models the height of the object thrown into the air is:
h(t) = -16t^2 + 99t, where the maximum height is between 90 and 100 m, the object starts on the ground, and the second zero is at least 10 seconds.