A rocket is shot into the air at a velocity of 48 ft/ sec. The equation is h(t) = -16t^2+48t+4
1) How far off the ground did the rocket start from?
2) Using the quadratic function, how high the rocket will be at its maximum point?
3) How long will it take the rocket to reach that maximum point?
4) Rewrite the function into vertex form.
5) Predict when the rocket will reach the ground.
6) Sketch the function. ( Label at least 5 points.)
h(t) = a t² + (t) = -16 t² + 48 t + 4
The coefficients are:
a = - 16 , b = 48 , c = 4
1)
When rocket start t = 0
h(0) = -16 ∙ 0² + 48 ∙ 0 + 4 = 4 ft
2)
The vertex of a quadratic equation a t² + b t + c is the highest or lowest point of that equation.
If a < 0 the vertex of a quadratic equation is the highest point of that equation.
If a > 0 the vertex of a quadratic equation is the lowest point of that equation.
In this case a = - 16 < 0 so the vertex of a quadratic equation is the highest point of that equation.
The t-coordinate of the vertex is:
t = - b / 2 a
So t = - 48 / 2 ∙ ( - 16 ) = - 16 ∙ 3 / - 16 ∙ 2 = 3 / 2
h(max) = h ( 3 / 2 ) = -16 ∙ ( 3 / 2)² + 48 ∙ 3 / 2 + 4 =
-16 ∙ 9 / 4 + 144 / 2 + 4 = - 36 + 72 + 4 = 40 ft
3 )
You already calculated t coordinate of vertex.
It will take 3 / 2 = 1.5 sec
4)
The vertex form of a quadratic equation:
h = a ( t - h )² + k
where
h = t coordinate of vertex = 3 / 2
k = h(max) = 40
a = - 16
So the vertex form of this quadratic equation:
h = a ( t - h )² + k
h = - 16 ( t - 3 / 2 )² + 40
h = - 16 ( t - 1.5 )² + 40
5)
The rocket will reach the ground when h(t) = 0
For that you must solve:
-16 t² + 48 t + 4 = 0
Try to do it.
The solutions are:
3 / 2 - √5/2 = - 0.081139
and
3 / 2 + √5/2 = 3.081139
Time can not be negative so t = 3.081139 sec
6)
Go on wolframalpha. c o m
In rectangle paste:
plot -16 t² + 48 t + 4
then click option =
#1. plug in t=0
#2. as with all quadratics, the vertex is at t = -b/2a = 3/2
#3. See #2
#4. h = -16(t^2 - 3t + 9/4) + 4 + 36 = _____
#5. solve h=0
1) To find how far off the ground the rocket started from, we need to find the initial height, which is the value of h(0).
Substitute t = 0 into the equation h(t) = -16t^2 + 48t + 4:
h(0) = -16(0)^2 + 48(0) + 4
h(0) = 0 + 0 + 4
h(0) = 4
Therefore, the rocket started 4 feet off the ground.
2) To find the maximum height of the rocket, we need to find the vertex of the quadratic function. The vertex form of a quadratic function is given by h(t) = a(t - h)^2 + k, where (h, k) is the vertex.
In the equation h(t) = -16t^2 + 48t + 4, the coefficient of t^2 is -16. Since a = -16, we can determine the vertex using the formula h = -b/2a where b = 48.
h = -48 / 2(-16)
h = -48 / -32
h = 1.5
Substitute h = 1.5 back into the equation to find the maximum height:
h(1.5) = -16(1.5)^2 + 48(1.5) + 4
h(1.5) = -16(2.25) + 72 + 4
h(1.5) = -36 + 72 + 4
h(1.5) = 40
Therefore, the rocket will reach a maximum height of 40 feet.
3) To find the time it takes for the rocket to reach the maximum height, we use the x-coordinate of the vertex, which is t = 1.5.
Therefore, it will take the rocket 1.5 seconds to reach its maximum height.
4) To rewrite the function h(t) = -16t^2 + 48t + 4 into vertex form, we complete the square:
h(t) = -16t^2 + 48t + 4
h(t) = -16(t^2 - (48/16)t) + 4
h(t) = -16(t^2 - 3t) + 4
h(t) = -16(t^2 - 3t + 9/4 - 9/4) + 4
h(t) = -16((t - 3/2)^2 - 9/4) + 4
h(t) = -16(t - 3/2)^2 + 36 + 4
h(t) = -16(t - 3/2)^2 + 40
Therefore, the function in vertex form is h(t) = -16(t - 3/2)^2 + 40.
5) To predict when the rocket will reach the ground, we need to find the value of t when h(t) = 0.
0 = -16t^2 + 48t + 4
16t^2 - 48t - 4 = 0
We can solve this quadratic equation using factoring, the quadratic formula, or completing the square. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
t = (-(-48) ± √((-48)^2 - 4(16)(-4))) / (2(16))
t = (48 ± √(2304 + 256)) / 32
t = (48 ± √2560) / 32
t = (48 ± 16√10) / 32
Therefore, the rocket will reach the ground at approximately t = 0.1 seconds or t = 3.9 seconds.
6) We can sketch the function h(t) = -16t^2 + 48t + 4 by plotting at least 5 points.
Using the vertex form h(t) = -16(t - 3/2)^2 + 40, we know that the vertex is at (3/2, 40). We can choose other values of t to find corresponding values of h(t):
t = 0: h(0) = -16(0 - 3/2)^2 + 40 = 4
t = 1: h(1) = -16(1 - 3/2)^2 + 40 = 36
t = 2: h(2) = -16(2 - 3/2)^2 + 40 = 40
t = 3: h(3) = -16(3 - 3/2)^2 + 40 = 36
t = 4: h(4) = -16(4 - 3/2)^2 + 40 = 4
Plotting these points, the graph of the function would be a downward-opening parabola with the vertex at (3/2, 40), passing through the points (0, 4), (1, 36), (2, 40), (3, 36), and (4, 4).
1) To find how far off the ground the rocket started from, we need to determine the initial height. In the given equation for the height of the rocket, h(t) = -16t^2 + 48t + 4, we can see that it is in the form of a quadratic equation, where the coefficient of t^2 is -16. Since the coefficient represents the acceleration due to gravity, and it is negative, we know that the equation represents a rocket launched from the ground.
The initial height can be found by evaluating h(0) since time, t, is zero at the start.
Plugging in t=0 in the equation, we have h(0) = -16(0)^2 + 48(0) + 4
Simplifying further, h(0) = 4
Therefore, the rocket started from 4 feet off the ground.
2) The maximum height of the rocket can be determined by finding the vertex of the quadratic equation. The vertex of a quadratic equation in the form h(t) = -16t^2 + 48t + 4 is given by the formula t = -b / (2a), where a and b are coefficients of t^2 and t respectively.
In this case, a = -16 and b = 48. Substituting these values into the formula, we get:
t = -(48) / (2(-16))
t = -48 / (-32)
t = 1.5
To find the maximum height, substitute this value of t back into the equation:
h(1.5) = -16(1.5)^2 + 48(1.5) + 4
h(1.5) = -16(2.25) + 72 + 4
h(1.5) = -36 + 76
h(1.5) = 40
Therefore, the rocket will reach a maximum height of 40 feet.
3) The time it takes for the rocket to reach the maximum point is given by the value of t we found in the previous step, which is 1.5 seconds.
4) To rewrite the function into vertex form, we can complete the square. The vertex form of a quadratic equation is given by h(t) = a(t - h)^2 + k, where (h, k) represents the coordinates of the vertex.
Starting with the equation h(t) = -16t^2 + 48t + 4, we can complete the square by factoring out the coefficient of t^2 (a = -16):
h(t) = -16(t^2 - 3t) + 4
Next, we need to find a value to add and subtract inside the parentheses to complete the square. Taking half of the coefficient of t (-3/2), we get -3/2.
h(t) = -16(t^2 - 3t + (-3/2) + 3/2) + 4
Now, we can rewrite the expression inside the parentheses as a perfect square trinomial:
h(t) = -16((t - 3/2)^2 - 9/4) + 4
Expand:
h(t) = -16(t - 3/2)^2 + 144/4 - 9/4 + 4
h(t) = -16(t - 3/2)^2 + 139/4
Therefore, the function h(t) can be rewritten in vertex form as h(t) = -16(t - 3/2)^2 + 139/4.
5) To predict when the rocket will reach the ground, we need to find the time at which h(t) = 0. In other words, we need to solve the equation -16t^2 + 48t + 4 = 0.
To solve this quadratic equation, we can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -16, b = 48, and c = 4.
Substituting the values into the quadratic formula, we get:
t = ( -48 ± √(48^2 - 4(-16)(4))) / (2(-16))
t = ( -48 ± √(2304 + 256)) / (-32)
t = ( -48 ± √(2560)) / (-32)
t = ( -48 ± 16√(10)) / (-32)
Simplifying further:
t = ( -3 ± √(10)) / 2
Since time cannot be negative in this case, we take the positive root:
t ≈ ( -3 + √(10)) / 2
Therefore, the rocket will reach the ground approximately at t ≈ -0.33 seconds.
6) To sketch the function h(t) = -16t^2 + 48t + 4, we can plot at least five points on the graph.
- Choose some t-values, plug them into the equation, and calculate the corresponding h(t) values. For example:
- When t = 0, h(0) = -16(0)^2 + 48(0) + 4 = 4 (starting point)
- When t = 1, h(1) = -16(1)^2 + 48(1) + 4 = 36
- When t = 2, h(2) = -16(2)^2 + 48(2) + 4 = 40
- When t = 3, h(3) = -16(3)^2 + 48(3) + 4 = 16
- When t = 4, h(4) = -16(4)^2 + 48(4) + 4 = -12
Using these points (0, 4), (1, 36), (2, 40), (3, 16), and (4, -12) as references, you can plot the graph of the function h(t) = -16t^2 + 48t + 4. The graph will have a parabolic shape opening downward.