An object launched straight puwar from the ground level wiht an initional velocity of 32 feet per second.The formula H=32T-16T^2
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9514 1404 393
Answer:
256 feet
Step-by-step explanation:
The vertex of the quadratic ax^2 +bx +c is found at x=-b/(2a). This means the time at which the rocket reaches it maximum height is ...
t = -128/(2(-16)) = 4 . . . . seconds
Evaluating h at t=4 gives the maximum height:
h = -16(4^2) +128(4) = -256 +512 = 256
The maximum height of the rocket is 256 feet.
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If you can read this value from "the graph at the left," you can save yourself some computation.
Looks like we have
which has divergence
By the divergence theorem, the integral of across
is equal to the integral of
over
, where
is the region enclosed by
. Of course,
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by attaching it to the disk
(call it
) so that
has boundary
.
Then by the divergence theorem,
Compute the integral in spherical coordinates, setting
so that the integral is
The integral of across
is equal to the integral of
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(without outward orientation, so that
Parameterize by
with and
. Take the normal vector to
to be
Then we have
Finally,