Respuesta :
a.) To find the maximum height, we can take the derivative of h(t). This will give us the rate at which the horse jumps (velocity) at time t.
h'(t) = -32t + 16
When the horse reaches its maximum height, its position on h(t) will be at the top of the parabola. The slope at this point will be zero because the line tangent to the peak of a parabola is a horizontal line. By setting h'(t) equal to 0, we can find the critical numbers which will be the maximum and minimum t values.
-32t + 16 = 0
-32t = -16
t = 0.5 seconds
b.) To find out if the horse can clear a fence that is 3.5 feet tall, we can plug 0.5 in for t in h(t) and solve for the maximum height.
h(0.5) = -16(0.5)^2 + 16(-0.5) = 4 feet
If 4 is the maximum height the horse can jump, then yes, it can clear a 3.5 foot tall fence.
c.) We know that the horse is in the air whenever h(t) is greater than 0.
-16t^2 + 16t = 0
-16t(t-1)=0
t = 0 and 1
So if the horse is on the ground at t = 0 and t = 1, then we know it was in the air for 1 second.
h'(t) = -32t + 16
When the horse reaches its maximum height, its position on h(t) will be at the top of the parabola. The slope at this point will be zero because the line tangent to the peak of a parabola is a horizontal line. By setting h'(t) equal to 0, we can find the critical numbers which will be the maximum and minimum t values.
-32t + 16 = 0
-32t = -16
t = 0.5 seconds
b.) To find out if the horse can clear a fence that is 3.5 feet tall, we can plug 0.5 in for t in h(t) and solve for the maximum height.
h(0.5) = -16(0.5)^2 + 16(-0.5) = 4 feet
If 4 is the maximum height the horse can jump, then yes, it can clear a 3.5 foot tall fence.
c.) We know that the horse is in the air whenever h(t) is greater than 0.
-16t^2 + 16t = 0
-16t(t-1)=0
t = 0 and 1
So if the horse is on the ground at t = 0 and t = 1, then we know it was in the air for 1 second.
The given function is a quadratic function and the graph of the quadratic
function is a parabola.
Correct response:
- a. 0.5 seconds
- b. Yes
- c. 1 second
Method by which the above responses are obtained:
Given parameters;
The given function is; h(t) = -16·t² + 16·t
Solution:
(a) The given function is a quadratic function of the form, y = a·x² + b·x + c;
[tex]\displaystyle The \ maximum \ height \ is \ at \ x = \mathbf{\frac{-b}{2 \cdot a}}[/tex]
Therefore, at the maximum height, we have;
[tex]\displaystyle t = \frac{-16}{2 \times (-16)} = \frac{1}{2} = 0.5[/tex]
- The maximum height is reached 0.5 seconds after the horse jumps.
(b) The maximum height reached by the horse is found as follows;
[tex]h_{max}[/tex] = h(0.5) = -16 × 0.5² + 16 × 0.5 = 4
The maximum height the horse can reach is 4 feet
Therefore;
- Yes, the horse can clear a fence that is 3.5 feet tall.
(c) The time the horse is in the air is given by the time it takes the horse to return to the ground, (h = 0 feet) as follows;
h(t) = 0 = -16·t² + 16·t = 16·t - 16·t²
Which gives;
0 = 16·t·(1 - t)
Which gives;
16·t = 0 ÷ (1 - t) = 0
t = 0
Dividing 0 by 16·t, we get;
(1 - t) = 0 ÷ 16·t = 0
1 - t = 0
1 = t
t = 1
Therefore, the horse is at the ground at t = 0 second, and t = 1 second
Which gives;
- The time the horse spends in the air is 1 second.
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