Question 118365: A fountain in the town square sprays water in a parabolic arc. The water spray starts at 1/2 metre above the ground and reaches a maximum height of 5 metres after 3 seconds. Determine the quadratic function that models the path followed by the water in the fountain and use it to determime the height of the water at 1&3/4 seconds . Round your answer to the nearest tenth of a metre .
Found 3 solutions by stanbon, ankor@dixie-net.com, bucky:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A fountain in the town square sprays water in a parabolic arc. The water spray starts at 1/2 metre above the ground and reaches a maximum height of 5 metres after 3 seconds. Determine the quadratic function that models the path followed by the water in the fountain and use it to determime the height of the water at 1&3/4 seconds . Round your answer to the nearest tenth of a metre .
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EQUATION Form:
h(t) = -9.81t^2+vot+so
Your equation:
h(t) = -9.81t^2+vot+(1/2)
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Max at (3,5) and max at -b/2a implies
-vo/(2*-9.81) = 3
vo = 6*9.81
vo = 58.86 m/s
Your eqution:
h(t) = -9.81t^2+58.86t+0.5
-----------
h(7/4) = -9.81(7/4)^2+58.86(7/4)+0.5
h(7/4) = 73.462 meters
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Cheers,
Stan H.
Answer by ankor@dixie-net.com(22740)  (Show Source):
You can put this solution on YOUR website! A fountain in the town square sprays water in a parabolic arc. The water spray starts at 1/2 metre above the ground and reaches a maximum height of 5 metres after 3 seconds. Determine the quadratic function that models the path followed by the water in the fountain and use it to determine the height of the water at 1&3/4 seconds . Round your answer to the nearest tenth of a metre.
:
Using the form: y = ax^2 + bx + c
:
y = .5 when x = 0; c = .5
:
Using the vertex, x = 3; y = 5
9a + 3b + .5 = 5
9a + 3b = 5 - .5
9a + 3b = 4.5
:
Using the symmetry of a parabola we know that 3 sec after the vertex, y = .5 again
x = 6; y = .5
36a + 6b + .5 = .5
36a + 6b = .5 - .5
36a + 6b = 0
:
Multiply the vertex equation by 2 and subtract from the above equation:
36a + 6b = 0
18a + 6b = 9
-------------subtracting eliminates b, find a
18a + 0b = -9
a = -9/18
a = -.5
:
Find b using 9a + 3b = 4.5
9(-.5) + 3b = 4.5
-4.5 + 3b = 4.5
3b = 4.5 + 4.5
3b = 9
b = 3
:
Our equation: y = -.5x^2 + 3x + .5
:
Illustrated by the graph:
:
Find the height after 1.75 seconds:
h = -.5(1.75^2) + 3(1.75) -.5
h = -1.53125 + 5.25 + .5
h = 4.2 meters after 1.75 seconds
Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website! A controversial problem ... to say the least. There is not a "real world" solution because,
according to the information given in the problem, the water cannot follow a gravitational
arc as predicted by Newtonian physics.
.
Stanbon used the ballistic equation from physics to "solve" the equation. He did almost everything
correctly except get a valid answer. Notice that when he put 7/4 into his equation for height
he got an answer of 73.462 meters for the height. But the problem tells you that the water
reaches a maximum height of 5 meters at t = 3 seconds. Also he forgot to divide the acceleration
of gravity (-9.8 m/s) by 2.
.
I initially used a similar approach. Starting with the ballistic equation from physics:
.
H(t) = (-9.8/2)t^2 + Vot + ho
.
I got that the initial height [ho] was 0.5 meters and substituted that. Then I substituted
5 for H(t) and 3 for t and solved for Vo ... finding Vo to be 16.2 m/s. But my answers were
out of whack also. I got an answer at 7/4 seconds that also exceeded the 5 meters that
is supposed to be the maximum height of the water.
.
ankor@dixie-net.com took an entirely different approach and cleverly developed an equation
that plots a path just as described in the problem. (Note: substitute t for x in his results
just to show that it is a relationship based on time.) His answer for the height of the stream
at t = 7/4 seconds is what you would logically expect ... 4.2 meters on its way up to the
peak of 5 meters which occurs at t = 3 seconds. Although it is mathematically correct, this
solution does not reflect what happens in the "real world." In reality there is no initial
velocity that you could apply to the water such that it would rise 4.5 meters (to a height
of 5 meters above the ground) in 3 seconds before falling back to its starting level of 0.5
meters.
.
So ... to whoever posted the question ... be aware of the "trick" in this problem. And
to whoever posed the problem ... time for you to remember classical physics.
.
Hope this helps explain the disparity of the two answers ...
.
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