Quadratic Relations

Question

Hardeep shoots a firecracker up off the roof of a building, and it falls to the ground below. The height of the firecracker can be modelled by the equation "h = -5.25t^2 + 42t + 12" where h is the height above the ground, in metres, and t is the elapsed time, in seconds.

a) How tall is the building?

b) What is the maximum height of the firecracker? When does it reach the maximum height?

c) How long does it take for the firecracker to hit the ground? Round your answer to the nearest hundredth.

Answer 1

h=-5.25t^2 +42t +12

the building is 12 meters tall

max height is when h'=0

-10.5t +42 =0

t = 42/10.5 = 4 seconds

max height = -5.25(4)^2 +42(4)+12

= 84 +168+12

= 96 meters high

it hits the ground when h=0

-5.25t^2 +42t +12 = 0

-t^2+8t -12/5.25 = 0

t^2 -8t = -12/21/4= -48/21

t^2-8t +16 = -48/21+16 = -48/21+336/21 = 288/21

(t-4)^2 = about 16.97

t-4= about +/- 4.12

t= about 4+/- 4.12

t = about -0.12 or 8.12 seconds to reach the ground

roughly twice the time to reach max height (2x4=8)

but since it started at 12 not 0, it takes another 0.12 seconds to reach the ground for a total 8.12 seconds