Larry likes skeet shooting, where a clay disc is shot into the air, and the participant tries to shoot it as it flies through the air. The discs are released from a firing mechanism that sits at ground level and shoots the disc on average a horizontal distance of 120 m on a parabolic path. On avergae, the skeet hits a height of 37.5 m after travelling a horizontal distance of 45 m.
a. Determine the equation of a function, in standard form, that represents the flight of the clay disc. Leave valuses in fractions.
b. Determine the average maximum height of the skeet by completing the square.
6 years ago
Answered By Kazi A
So the key to answering this question is understanding that if you have any three points on a parabola, you can find its equation (and hence find any other point on the parabola). In general, the equation for a parabola is given by the quadratic equation:
$y=ax^2+bx+c$y=ax2+bx+c .... (i)
For our problem, I have sketched a graph. We choose the origin to be location of the gun that fires the disc, making x=0 when y=0. Thus we already have one point, (0,0) as part of the problem. Plugging x = 0 and y = 0 into the equation (i) above, we get:
$0=a\left(0\right)^2+b\left(0\right)+c$0=a(0)2+b(0)+c
$\therefore c=0$∴c=0
We are also told that the total horizontal distance the disc flies is 120m. In other words, the disc falls to the ground, where y=0, when x=120. This gives us another pair of points to plug into equation (i), along with c=0:
$a\left(120\right)^2+b120=0$a(120)2+b120=0 ...(ii)
This gives us one equation with two unknowns: a and b. You can similarly get one more equation with a and b if you plug in the last pair of values of x,y that we are given: when x=45, y=37.5. Once you have these two equations, all you have to do is solve them as a pair of simultaneous equations, and then you will have the whole quadratic equation, with the values of a,b and c. To calculate the maximum height, simply plug in the value of x where this maximum height occurs, which would be exactly halfway to the maximum distance, since parabolas are perfectly symmetric. This is much simpler than completing the square.
Attached Graph: