Trajectory of a Bow and Arrow

Bows and arrows have a great deal of variance to them. From the type of bow you use, the material, the bowstring, the wind, and where you’re shooting from, many different things could happen.

Cool guy is in an archery competition and he is about to take his first shot. Cool guy releases the arrow from a height of 3 feet and it hits its highest point (5 feet) when the arrow is 50 feet from where it was released. The target is 120 feet away and the center of the bulls eye is exactly 3 feet above the ground, and 4 inches in diameter. Assuming that Cool guy shoots a perfectly straight shot, will he hit a bulls eye?

We need to use the vertex form of the equation y = a(x – h)^2 + k To find the vertical height of the arrow as it approaches the target. Since the arrow has a maximum height of 5feet when it has traveled 50 feet toward the target, we know that the vertex is at (50, 5). Substituting this into the equation, we get: y = a(x – 50)^2 + 5

To find the stretching factor, a, we need to plug in a point other than the vertex. Since the arrow was initially launched at a height of 3 feet, the parabola goes through the point (0,3). Replacing x and y in the equation, we get:3 = a(0 – 50)^2 + 53 = 2500a + 5-2 = 2500aa = -0.0008
This gives a final equation of y = -0.0008(x – 50)^2 + 5We can find the height of the arrow when it hits the target by plugging in 120 feet in for x.y = -0.0008(120 – 50)^2 + 5y = 4.888 feet

The bulls eye is at 3 feet above the ground. Subtracting the height from 3, we will find the place on the target that the arrow hits.3 – 4.888 = -1.888 feet below the bulls eye-1.888 ft * 12 in/ft = 22.656 inches below the bulls eye
Cool guy missed the bulls eye. In fact he missed the whole target. Maybe next time, Cool Guy.