Using his driver on the 7th tee, Chris hits an excellent shot, right down the middle of the fairway. The ball follows the parabolic path shown in the figure, described by the quadratic function: h(x)=0.5x-0.002x^2. This relates the height, h, of the ball above the ground to the distance x it has traveled down the course. Distances are measures in feet.
a) How high is the ball after it has traveled 20 feet horizontally down the course?
b) Factor the quadratic function that describes the ball’s path. Notice that x = 0 when x= 0. What is the physical meaning of this?
c) How far away from the tee does the ball hit the ground?
d) What is the ball’s maximum height above the ground and what distance x, from the tee, does this occur at?
Could you please teach me these problems..?
Sure, I'd be happy to help you with these problems! Let's go through them one by one:
a) To find out how high the ball is after it has traveled 20 feet horizontally down the course, we need to plug in x = 20 into the quadratic function h(x) = 0.5x - 0.002x^2.
h(20) = 0.5(20) - 0.002(20)^2
= 10 - 0.002(400)
= 10 - 0.8
= 9.2 feet
Therefore, the ball is 9.2 feet above the ground after it has traveled 20 feet horizontally down the course.
b) To factor the quadratic function h(x) = 0.5x - 0.002x^2, we can rewrite it as:
h(x) = -0.002x^2 + 0.5x
Factoring out the common factor -0.002, we have:
h(x) = -0.002(x^2 - 250x)
Now we can factor the quadratic expression (x^2 - 250x). The factored form is:
h(x) = -0.002x(x - 250)
The physical meaning of x = 0 is that it represents the position of the ball at the starting point, which is the tee. When x = 0, the ball has not traveled any distance horizontally down the course yet.
c) To find out how far away from the tee the ball hits the ground, we need to find the x-coordinate when the height, h(x), becomes zero. In other words, we need to solve the equation:
0.5x - 0.002x^2 = 0
We can factor out a common factor of x:
x(0.5 - 0.002x) = 0
Setting each factor equal to zero, we have:
x = 0 or 0.5 - 0.002x = 0
The first solution, x = 0, is the starting point at the tee. For the second solution:
0.5 - 0.002x = 0
0.002x = 0.5
x = 0.5/0.002
x ≈ 250
Therefore, the ball hits the ground approximately 250 feet away from the tee.
d) The maximum height of the ball above the ground occurs at the vertex of the parabolic path. The x-coordinate of the vertex can be found using the formula x = -b / (2a), where the quadratic function is in the form ax^2 + bx + c.
Here, a = -0.002 and b = 0.5. Plugging these values into the formula, we get:
x = -0.5 / (2*(-0.002))
x = -0.5 / (-0.004)
x ≈ 125
Therefore, the ball reaches its maximum height of approximately 125 feet above the ground at a distance of 125 feet from the tee.
I hope this helps! Let me know if you have any further questions.
a) To find the height of the ball after it has traveled 20 feet horizontally down the course, we substitute x=20 into the quadratic function h(x)=0.5x-0.002x^2:
h(20) = 0.5(20) - 0.002(20)^2
= 10 - 0.002(400)
= 10 - 0.8
= 9.2 feet
Therefore, the ball is 9.2 feet above the ground after traveling 20 feet horizontally down the course.
b) To factor the quadratic function h(x)=0.5x-0.002x^2, we first set it equal to zero to find the x-intercepts:
0.5x - 0.002x^2 = 0
Now, we can factor out x:
x(0.5 - 0.002x) = 0
From this equation, we can see that one of the roots is x=0. This means that when x=0 (at the start of the tee), the ball is at the ground level. The physical meaning of this is that the ball is initially hitting the ground from the tee.
c) To find the distance from the tee where the ball hits the ground, we need to find the x-value when the height h(x) equals zero:
0.5x - 0.002x^2 = 0
Factoring out x, we get:
x(0.5 - 0.002x) = 0
So either x=0 or 0.5 - 0.002x = 0.
Solving the second equation for x:
0.5 - 0.002x = 0
0.002x = 0.5
x = 0.5/0.002
x = 250
Therefore, the ball hits the ground 250 feet away from the tee.
d) To find the maximum height of the ball above the ground, we can complete the square of the quadratic function h(x)=0.5x-0.002x^2:
h(x) = -0.002x^2 + 0.5x
To complete the square, we need to divide the coefficient of x by 2 and square it:
h(x) = -0.002(x^2 - 250x + (250/0.002)^2) + (250/0.002)^2
Simplifying this, we get:
h(x) = -0.002(x - 125)^2 + 62500
Therefore, the maximum height occurs when the term inside the parentheses is zero, which happens when x=125. Substituting this value into the equation, we get:
h(125) = -0.002(125 - 125)^2 + 62500
= 62500
Therefore, the ball's maximum height above the ground is 62500 feet, and this occurs at a distance of 125 feet from the tee.
a) To find the height of the ball after it has traveled 20 feet horizontally down the course, we need to substitute x = 20 into the quadratic function h(x) = 0.5x - 0.002x^2.
h(20) = 0.5(20) - 0.002(20)^2
= 10 - 0.002(400)
= 10 - 0.002(400)
= 10 - 0.8
= 9.2
Therefore, the ball is 9.2 feet high after it has traveled 20 feet horizontally down the course.
b) To factor the quadratic function h(x) = 0.5x - 0.002x^2, we can set it equal to zero and solve for x:
0.5x - 0.002x^2 = 0
To factor this quadratic equation, we can divide both sides by x:
0.5 - 0.002x = 0
Rearranging the equation, we get:
0.002x = 0.5
Dividing both sides by 0.002, we find:
x = 250
The physical meaning of x = 0 is that it represents the starting point, the tee. In this context, it means that when x = 0, the ball has not yet been hit and is still at ground level.
c) To find the distance from the tee where the ball hits the ground, we need to find the x-coordinate where h(x) = 0. Setting h(x) = 0 and solving for x:
0.5x - 0.002x^2 = 0
We can factor out x:
x(0.5 - 0.002x) = 0
This equation will be satisfied when either x = 0 or 0.5 - 0.002x = 0.
From the previous step, we already know that x = 0 represents the starting point at the tee.
To find the other solution, we can solve:
0.5 - 0.002x = 0
0.002x = 0.5
Dividing both sides by 0.002, we find:
x = 250
Therefore, the ball hits the ground 250 feet away from the tee.
d) To find the ball's maximum height above the ground and the distance x from the tee, we need to determine the vertex of the parabolic function h(x) = 0.5x - 0.002x^2.
The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a = -0.002 and b = 0.5:
x = -0.5 / (2 * (-0.002))
x = -0.5 / -0.004
x = 125
So, the ball reaches its maximum height 125 feet away from the tee.
To find the maximum height, we substitute x = 125 into the equation:
h(125) = 0.5(125) - 0.002(125)^2
= 62.5 - 0.002(15625)
= 62.5 - 31.25
= 31.25
Therefore, the ball's maximum height above the ground is 31.25 feet, and it occurs at a distance of 125 feet from the tee.
h(x)=0.5x-0.002x^2
h(x)=.5x(1-.004x^2)=.5/.004 * (x*(250-x^2)
when x=0, x=15.81 (ie, sqrt250), height is zero.
max height occurs at 15.81/2 feet downrange, figure that out from h(x)