SOLUTION
(a)
$h=-16(3)^2+48(3)+50$
$=50$ ft
(b)
Since when $t=0, h=50$, and when $t=3, h=50$ also,
by symmetry the highest point is reached half way in between, so $t=(0+3)/2=1.5$s
Therefore the maximum height is
$h=-16(1.5)^2+48(1.5)+50$
$=86$ ft
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Saturday, 5 December 2015
Quadratic function. An object is thrown upward from a height of $50$ feet with an initial velocity of $48$ ft/s.
QUESTION
SOLUTION
(a)
$h=-16(3)^2+48(3)+50$
$=50$ ft
(b)
Since when $t=0, h=50$, and when $t=3, h=50$ also,
by symmetry the highest point is reached half way in between, so $t=(0+3)/2=1.5$s
Therefore the maximum height is
$h=-16(1.5)^2+48(1.5)+50$
$=86$ ft
SOLUTION
(a)
$h=-16(3)^2+48(3)+50$
$=50$ ft
(b)
Since when $t=0, h=50$, and when $t=3, h=50$ also,
by symmetry the highest point is reached half way in between, so $t=(0+3)/2=1.5$s
Therefore the maximum height is
$h=-16(1.5)^2+48(1.5)+50$
$=86$ ft
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