footballer kicks a ball that follows a a footballer kicks a ball that follows a parabolic trajectory. the ball just clears a 25m high pole at the top of its trajectory. the pole is 30m fron the footballer. find the equation of the trajectory in turning point form. y=a(x-h)squared +k
not a very satisfying answer, left in so complicated a form.
Also, "just clears" does not necessarily mean that the crossbar is at the vertex of the parabola. The ball could still be rising, or back on its way down.
So we only have two points on the parabola: (0,0) and (30,25).
Some other constraint is needed to pin down the equation.
But, if the crossbar is in fact at the vertex, then we clearly have
y = ax(x-60)
setting y(30)=25, we have a*30*(-30) = 25, so a = -1/36
y = -1/36 (x-30)^2 + 25
Thank you for your feedback! You're correct that the interpretation of "just clears" could vary, and that the question could benefit from a more precise description.
Using the assumption that the crossbar is at the vertex, your answer is correct and provides a simpler expression for the equation. Thank you for sharing this alternative method!
To find the equation of the trajectory of the ball in turning point form (y = a(x-h)^2 + k), we need to determine the values of a, h, and k.
Given information:
- The ball just clears a 25m high pole at the top of its trajectory.
- The pole is 30m from the footballer.
Let's proceed step-by-step:
Step 1: Let's determine the turning point (h, k) of the parabolic trajectory.
- Since the ball just clears the 25m high pole at the top of its trajectory, the maximum height of the ball is 25m.
- The pole is located 30m from the footballer, so the x-coordinate of the turning point is h = 30m.
Therefore, the turning point is (30, 25).
Step 2: Let's determine the value of a.
- The value of a determines the steepness of the parabola. To find it, we need one more point on the trajectory.
- Since the turning point is at the highest point of the trajectory, the ball must touch the ground at another point on the same level.
- The horizontal distance from the footballer to the ground at the same level is 30m, because at that distance, the height of the pole is already 0.
- So, another point on the trajectory is (60, 0).
To find a, substitute the values of (30, 25) and (60, 0) into the equation in turning point form:
25 = a(30 - 30)^2 + k
0 = a(60 - 30)^2 + k
Simplifying these equations:
25 = a(0)^2 + k
0 = a(30)^2 + k
25 = k
900a = -k
Since k = 25, we can substitute it into the third equation:
900a = -25
Solving for a:
a = -25 / 900
a = -1/36
Step 3: Now that we have the values of a, h, and k, we can write the equation of the trajectory in turning point form:
y = a(x - h)^2 + k
Substituting the known values:
y = (-1/36)(x - 30)^2 + 25
Therefore, the equation of the trajectory in turning point form is:
y = (-1/36)(x - 30)^2 + 25
We can start by using the vertex form of a parabola, which is y = a(x-h)^2 + k, where (h,k) is the vertex of the parabola.
Since the ball just clears a 25m high pole, the highest point of the trajectory will be 25m above the ground. Let's call this point (x_max, 25).
To find x_max, we need to consider the symmetry of the parabola. The ball reaches its maximum height at the midpoint of its horizontal range, which is halfway between the footballer and the pole. So x_max = (30 + 0)/2 = 15.
Now we have one point on the parabola: (15, 25).
To find the coefficient a, we can use the fact that the parabola passes through the point (0,0). Plugging x=0 and y=0 into the equation gives:
0 = a(0-h)^2 + k
0 = ah^2 + k
Now we have two equations with two unknowns:
a(15-h)^2 + k = 25
ah^2 + k = 0
To eliminate k, we can subtract the second equation from the first:
a(15-h)^2 - ah^2 = 25
Simplifying, we get:
a(225 - 30h + h^2) - ah^2 = 25
a(225 - 30h) = 25
a = 25 / (225 - 30h)
Now we can substitute this expression for a into one of the earlier equations to solve for h:
ah^2 + k = 0
25h^2 / (225 - 30h) + k = 0
k = -25h^2 / (225 - 30h)
Finally, we have the equation of the parabola in vertex form:
y = a(x-h)^2 + k
y = [25 / (225 - 30h)](x - 15)^2 - 25h^2 / (225 - 30h)
To write this in turning point form, we need to complete the square:
y = [25 / (225 - 30h)](x^2 - 30x + 225) - 25h^2 / (225 - 30h)
y = [25 / (225 - 30h)](x^2 - 30x + 225 - 225 + 25h^2)
y = [25 / (225 - 30h)](x - 15)^2 + 25(1 - h^2 / (225 - 30h))
So the equation of the trajectory in turning point form is:
y = [25 / (225 - 30h)](x - 15)^2 + 25(1 - h^2 / (225 - 30h))
where h is a solution of the equation ah^2 + k = 0, with a = 25 / (225 - 30h) and k = -25h^2 / (225 - 30h).