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.