You can
put this solution on YOUR website!The path of a falling object is given by the function s=-16t^2+v0t+s0 where v0 represents the initial velocity in ft/sec and s0 represents the initial height. The variable t is time in seconds, and the s is the height of the object in feet.
a) If a rock is thrown upward with an initial velocity of 32 feet per second from the top of a 40-foot building, write the height equation using this information.
s(t)=-16t^2+32t+40
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b) How high is the rock after 0.5 seconds?
s(0.5) = -16(0.5)^2+32(0.5)+40 = 50 ft.
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c) After how many seconds will the rock reach maximum height?
Maximum occurs when t = -b/2a = -32/(2*-16) = 1 second
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d)What is the maximum height?
Maximum height occurs at s(1) = -16+32+40 = 56
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Cheers,
Stan H.
You can
put this solution on YOUR website!
a. You are given that
and
, so just plug in the values:
b. You need to evaluate
So the height (measured from the ground) after one-half second is 52 feet.
c. This is a quadratic equation and if you graphed it on a coordinate plane with s as the vertical axis and t as the horizontal axis, you would have a convex down parabola. The maximum height will be reached at time equal to the value of t at the vertex of the parabola.
The vertex of any parabola described by
is located at
(
,
)
For this problem,
and
, hence
second.
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You can also do this part with the calculus
A local minimum or maximum is found where the first derivative equals 0.
For this problem, s'(t)=
, so if
then
To determine if this is a maximum or minimum, evaluate the second derivative at the same value
s"(t)=
. Since the second derivitive is less than zero, this is a maximum.
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d. The actual maximum height is just the function evaluated at the time for maximum height, i.e. 1 second, or
And the maximum height is 56 feet.
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