SOLUTION: Equation h(t)=-1/2(9.8)x^2+40x+2
When does the arrow almost instantaneously stop in the air? How high is the arrow at that time? What happens on the graph of this function to repr
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Question 813382: Equation h(t)=-1/2(9.8)x^2+40x+2
When does the arrow almost instantaneously stop in the air? How high is the arrow at that time? What happens on the graph of this function to represent this situation?
When does the arrow strike the ground? Explain how you determine this value. What part of the parabola represent this situation? How could this be set up as an equation
Found 2 solutions by richwmiller, TimothyLamb:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
a) At its maximum or vertex. t=4.08163
b) 83.6327
c) It is the vertex- the highest point
vertex | (4.08163, 83.6327)
d) x = 8.21296
e) when y=0 ;when the parabola cross the x axis
set h(t)=0 and solve for x
f)when the parabola cross the x axis
g) set h(t)=0 and solve for x
Answer by TimothyLamb(4379) (Show Source): You can put this solution on YOUR website!
the given quadratic function:
h(t) = -1/2(9.8)x^2 + 40x + 2
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is represented by the quadratic equation:
-4.9t^2 + 40t + 2 = 0
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the above quadratic equation is in standard form, with a=-4.9, b=40, and c=2
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to find the maximum height reached by the arrow, plug this:
-4.9 40 2
into this: https://sooeet.com/math/quadratic-equation-solver.php
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the maximum point of the quadratic equation is: ( t= 4.08163265, h(t)= 83.6326531 )
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Answer 1:
When does the arrow almost instantaneously stop in the air? at time = 4.08163265 seconds
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Answer 2:
How high is the arrow at that time? height = 83.6326531 meters
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(NOTE: the question doesn't give the units of height or of time, but based on the 9.8 value used in the equation, which is the acceleration due to gravity at the Earth's surface in meters/sec^2, the height units are meters, and the units of time are seconds.)
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Answer 3:
What happens on the graph of this function to represent this situation?
the parabola that describes the motion of the arrow is at a maximum, and no value of h(t) is larger for any time t.
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Answer 4:
When does the arrow strike the ground?
from the above solver we also get the roots of the quadratic equation, which are: -0.0496974451 and 8.21296275
these are the t-intercepts of the quadratic equation, the t values for which h(t) = 0.
the negative root doesn't make sense because real time is never negative.
the positive root gives the time that the arrow strikes the ground, which is 8.21296275 seconds
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Answer 5:
What part of the parabola represent this situation?
already explained above, h(t) = 0 at the positive root, which occurs at 8.21296275 seconds
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Answer 5:
How could this be set up as an equation?
already answered above:
the given quadratic function [ h(t) = -1/2(9.8)x^2 + 40x + 2 ] is represented by the quadratic equation:
-4.9t^2 + 40t + 2 = 0
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Solve and graph linear equations:
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https://sooeet.com/math/linear-regression.php
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