The function h(t) = -16t^2 + 32x + 240 depicts the flight of a canon ball shot from 240 feet in the air with an initial velocity of 32 feet per second. 2.How high does the canon ball fly

Answer 1

h(t) = -16t² + 32t + 240

t = time (in seconds) where the canon ball launched

h(t) = height (in feet)

This question asks about the maximum height from where the canon ball is launched in the air from its initial height of 240 feet.

1.) Factor the quadratic equation using greatest common factor (GCF = 16), then diamond or box method to find two solutions.

-16t² + 32t + 240 = -16(t² - 2t - 15) = -16(t² - 5t + 3t - 15) = -16[t(t - 5) + 3(t - 5)] = -16(t - 5)(t + 3)

Apply the zero product property to get t = 5 or t = -3.

2.) Add the two solutions and divide it by 2 to get (5 + (-3)) / 2 = 2/2 = 1.

3.) Plug in x = 1 to the quadratic equation to find the maximum height: -16(1 - 5)(1 + 3) = -16(-4)(4) = 256

The canon ball will fly at 256 feet after 1 second. This represents the vertex at (1, 256).