A firework is launched upward to celebrate the end of the school year. The height of the firework, in metres, above the ground after t seconds, is modelled by the equation: h = -4.9t2 + 49t + 1.5 a) What is the height of the firework at 3 seconds? /2 b) /4 How long does it take for the firework to hit the ground? c) What is the maximum height reached by the firework? /4 d) What is the y-intercept and what does it tell you in the context of this word problem? /2

Given a firework's height is described by h = -4.9t² +4.9t +1.5, you want the height at 3 seconds, the time it hits the ground, the maximum height, and the value and meaning of the y-intercept.

a) 3 seconds

The height at 3 seconds is found by evaluating the formula using t=3.

(-4.9(3) +49)·3 +1.5 = 104.4

The height of the firework at 3 seconds is 104.4 meters.

b) Ground

We can write the equation in vertex form to help find both the maximum height and the zeros.

h = -4.9(t² -10t) +1.5 = -4.9(t² -10t +25) +1.5 +4.9(25)

h = -4.9(t -5)² +124

For h = 0, we have ...

-4.9(t -5)² = -124 . . . . . . . . . . . . . set to 0, subtract 124

t -5 = √(124/4.9) ≈ 5.0305 . . . . . divide by -4.9, take square root

t = 10.0305 . . . . . . . add 5

It takes about 10.03 seconds for the firework to hit the ground.

c) Height

The vertex (h, k) can be read from the vertex form y = a(x -h)² +k. It is ...

(t, h) = (5, 124)

The maximum height reached is 124 meters.

d) Y-intercept

There is no "y" in the equation. We assume the intent of "y-intercept" is the value of h when t=0, the point where the graph joins the vertical axis. That value is the constant in the equation, 1.5 (meters).

The y-intercept is the value of h when t=0, the initial height of the firework. It is 1.5 meters.