Question

The equation that models the height of an aerial bomb t

seconds after it is shot upwards from the ground with an initial
velocity of 80 ft/sec is given by h = -167 + 80t What is the
maximum height above ground level that the aerial bomb will
reach?
а
180 feet
b 120 feet
C 100 feet
d 10 feet

Answer 1

C. 100 feet

Explanation:

The equation can be written in vertex form as ...

h = -16(t^2 -5t) = -16(t^2 -5t +6.25) +16(6.25) = -16(t -2.5)^2 +100

The vertex of the height curve is (2.5, 100), so the bomb reaches a maximum height of 100 feet after 2.5 seconds.