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Parabola of a Horse's Jump

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Sawyer Panara

on 28 May 2013

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Transcript of Parabola of a Horse's Jump

Presented by Sawyer Panara The Parabola of a
Horse's Jump There are multiple ways math is used in the equine world. Feeding
A horse can only consume 2.5% of his weight each day.
Ratios are used when mixing nutrients to make sure all the requirements are met.
Estimating a horse's weight is a mathematical equation.
(G^2 x L)/12,000 Arenas
A dressage arena for competition must be 20 by 60 meters, Pythagoras's Theorem can be used to make sure the length and angles are correct. Jump Spacing
In showjumping, the number of strides determine the spacing between each jump. When you put it together you can start to see a curve... kind of like a parabola! I chose to focus specifically on a horse's jump and how it creates a parabola. But before I get into that, you need to know the basics of jumping. In a proper jump...
the take off and landing are the same distance from the fence - a, e
the peak is directly over the highest point of the fence - c Now put into motion... Vertex X-intercept X-intercept The Parabola in Relation to the Jump y = ax^2 + bx + c peak (mid flight) represents the vertex
take off and landing points represent the x-intercepts
the relationship is always negative, or minimum
axis of symmetry goes vertically through the highest point of jump (not necessarily center)
the equation represents the path of the horses FEET axis of symmetry Notice that the rider is bareback and bridle-less!...awesome(: I took multiple photos of horses jumping and put each one on a graph. In each photo the horse is at a different stage of the jump, so each one gave me a different piece of information. common piece of info gathered from all pictures: axis of symmetry.
in all of the photos, fence was visible
if I know highest point of fence, I know exactly where the axis of symmetry should be if I knew take off, I knew landing
with that single piece of info I was able to find both x-intercepts in order to clear jump, parabola's vertex had to be greater than highest point of fence
when vertex unknown, made an inequality to show where vertex must be for a successful jump Info From the Photos Thanks! That's all... AOS: x=2
X-intercepts: (-4,0) , (8,0)
Vertex: > (2,3.5) AOS: x=2
X-intercepts: (-4.5,0) , (7.5,0)
Vertex: > (2,5) AOS: x=3
X-intercepts: unknown
Vertex: (3,7) AOS: x=6
X-intercepts: (-3,0) , (15,0)
Vertex: > (6,3) AOS: x= -6
X-intercepts: (4,0) , (-10,0)
Vertex: > (-6,3.5)
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