2. 3 part Jump
• The - ApproachIncludesthe three strides before a
fence and the 'takeoff' in front of the fence.
• The Jump- The air time as the horse rises in the
air, experiences 'hangtime', and begins to descend
back down to the ground.
• The Landing- Includes actual 'impact' of the horse
hitting the ground, and the follow through as horse
and rider continue on to the next fence.
4. The Approach
• The horse and rider possess only Kinetic
Energy (if the ground is considered to be 'y-
zero’). This kinetic energy is expressed
by: KE = 1/2mv2
5. The Jump
• Over the fence - Horse and rider reach
maximum height and their velocity is reduced
to zero; they possess only Potential
Energy, PE. This Potential Energy is expressed
by: PE = mgy
6. The Landing
• The horse returns to the ground, horse and
rider are once again at 'y-zero' and possess
only Kinetic Energy, KE, expressed
by: KEf= 1/2mvf
2
7. What if….
If the fence is y= 5 1.52 m (5 ft).Incalculation,assume
the horse may jump up to six inches higher than the
fence (depending on their perception) making y =
1.52m + 0.15m or y=1.52m
Due to non conservative forces, such as air resistance
and heat, the potential energy reached at the top of
the jump will only be about 80% of the kinetic energy
present on approach.
1/2v2 =
0.8(9.8m/s2*1.52m) vlo
w = 6.1m/s vhigh =
6.4m/s
8. Projectile motion
Range equation: R = (vo
2/g) (sin
2θ)
Height equation: H = (vo
2 sin2θ)/(2g)
For this analysis we are going to determine the
initial velocity at which the horse must leave
the ground in order to get over a jump of a
certain height and width.
9. With what initial velocity will a horse
need to take off in order to jump a
1.5m high fence?
Assume the horse leaves the ground at
45°. (angle would change depending on the
different properties of the intended jump)
H = (vo
2 sin2θ)/(2g)
vo= √((H*2g)/(sin2θ))
vo= √((1.5m*2*9.8m/s)/(sin245)vo= 7.7m/s
10. Horse power
The term horsepower was coined by James Watt, the
inventor of the steam engine, when he needed a unit
of power large enough to describe the output of his
new invention. He worked in a coal mine and observed
the work of the ponies that were used in the mine to
haul coal. He saw that a mine pony could do about
22,000 foot-pounds of work in a minute. He then
increased that number by 50% to make the
measurement of horsepower (not ponypower!) at
33,000 foot-pounds of work in one
minute. So, according to Watt, a horse can haul 330
pounds of coal 100 feet in one minute.
