[EQUATION PROVIDED THIS TIME] What is the maximum height the fireworks reach? How high will the fireworks be 1 second after launch? How long after launch will the fireworks explode?

Question

For a 4th of July celebration, some fireworks are launched off of a hill 80 feet above a lake. The fireworks will explode after reaching their maximum height before they fall down into the lake below. The fireworks' height above the surface of the lake is represented by the equation shown below. What is the maximum height the fireworks reach?

h(t)=-16t²+64t+80

How long will it take the fireworks to fall into the lake? What is the maximum height the fireworks reach? How high will the fireworks be 1 second after launch? How long after launch will the fireworks explode?

Answer 1

This equation represents a downward facing parabola. When the fireworks fall into the lake height will be 0 so solve the equation -16t^2 +64t +80=0. Start by diving both sides of the equation by -16 to make your life easier this gives you t^2+4t--5=0

(t-5)(t+1)=0 t=5 t=-1. The answer of -1 can be discarded as this would require you to go back in time. So it takes 5 seconds for the fireworks to fall into the lake

If you want to know the height after 1 second just substitute 1 for t so height is -16 (1)^2 +64(1) +80=128 FT

The max height will occur at the vertex of the parabola/remember the the x coordinate of the vertex is at -b/2a which is -64(-2x16) =2

Now substitute 2 for t and calculate the height -16(4) +64(2) +80 = 144 ft

Answer 2

The height is zero at the lake.

-16t²+64t+80 = 0

t² - 4t - 5 = 0

(t + 1)(t - 5) = 0

t + 1 = 0

t = -1

Since we can't go back in time, we disregard this solution.

t - 5 = 0

t = 5 seconds

h(1) = -16(1)² + 64(1) + 80 = -16 + 64 + 80 = 128 ft

x = -64/2(-16) = 2

h(2) = -16(2)² + 64(2) + 80 = -64 + 128 + 80 = 144 ft

Answer 3

This is a quadratic equation because of the "x²" term. A quadratic is of the form y=ax²+bx+c. The maximum height will be at the vertex. We know that the "x" value of the vertex is found by the equation x=-b/2a.

y=ax²+bx+c

h(t)=-16t²+64t=80

y=h(t)

a=-16

b=64

c=80

x=-b/2a

x=64/(2)(-16)

x=2

Therefore the max height is when t=2sec: h(2)=-16(2)²+64(2)+80

h_max=(-16)(4)+128+80

h_max=-64+128+80

h_max=144ft

1 second after launch the height is h(1)=-16(1)²+64(1)+80=128ft.

The time it hits the water is when h(t)=0. So, you must solve 0=-16t²+64t+80 for "t" by factoring.

0=-16(t²-4t-5)

0=-16(t-5)(t+1)

t=5 t=-1

Because time cannot be negative, our answer is 5 seconds.

Hope this helps!