OpenStudy (anonymous):

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

2 years ago
OpenStudy (mehek14):

what did you get?

2 years ago
OpenStudy (anonymous):

okay so, ik i can take a 4 out of all of them

2 years ago
OpenStudy (welshfella):

maximum height is when velocity = 0

2 years ago
OpenStudy (mehek14):

use the same formula as last time

2 years ago
OpenStudy (anonymous):

okay -45/2(-16)

2 years ago
OpenStudy (mehek14):

yes and then?

2 years ago
OpenStudy (anonymous):

-45/-32 1.4

2 years ago
OpenStudy (mehek14):

it can actually be rounded to 1.41

2 years ago
OpenStudy (mehek14):

now plug 1.41 into the equation

2 years ago
OpenStudy (anonymous):

okay then plug it in

2 years ago
OpenStudy (mehek14):

yes

2 years ago
OpenStudy (anonymous):

i got 98

2 years ago
OpenStudy (mehek14):

\(h(t) = -16*1.41^2 + 45*1.41 + 40\\ 1.41^2=1.9881*-16=-31.81\\45*1.41=63.45\\63.45-31.81=31.64\\31.64+40=71.64\) 71.64 can be rounded to 72

2 years ago
OpenStudy (anonymous):

ohh okay.

2 years ago
OpenStudy (anonymous):

thank you!

2 years ago
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