Question

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

Answer 1

what did you get?

Answer 2

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

Answer 3

maximum height is when velocity = 0

Answer 4

use the same formula as last time

Answer 5

okay
-45/2(-16)

Answer 6

yes and then?

Answer 7

-45/-32
1.4

Answer 8

it can actually be rounded to 1.41

Answer 9

now plug 1.41 into the equation

Answer 10

okay
then plug it in

Answer 11

yes

Answer 12

i got 98

Answer 13

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

Answer 14

ohh okay.

Answer 15

thank you!