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).
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
3 answers
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
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!
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!