An object is launched upward at 45 ft/sec from a platform that is 40 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t2 + 45t + 40? 65 feet 72 feet 80 feet 93 feet

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An object is launched upward at 45 ft/sec from a platform that is 40 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t2 + 45t + 40? 65 feet 72 feet 80 feet 93 feet

Mathematics
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what did you get?
okay so, ik i can take a 4 out of all of them
maximum height is when velocity = 0

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Other answers:

use the same formula as last time
okay -45/2(-16)
yes and then?
-45/-32 1.4
it can actually be rounded to 1.41
now plug 1.41 into the equation
okay then plug it in
yes
i got 98
\(h(t) = -16*1.41^2 + 45*1.41 + 40\\ 1.41^2=1.9881*-16=-31.81\\45*1.41=63.45\\63.45-31.81=31.64\\31.64+40=71.64\) 71.64 can be rounded to 72
ohh okay.
thank you!

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