A ball is thrown in the air from a ledge. It's height in feet represented by f(x)=16(x^2-6x-7), where x is the number of seconds since the ball has been thrown. The height of the ball is 0 feet when it hits the ground. How many seconds does it take the ball to reach the ground?

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Answer:

Step-by-step explanation:

Since we know that the height is 0, we can figure out how long it took the ball to reach the ground by setting [tex]f(x) = 0[/tex] and solving for [tex]x[/tex]:

[tex]f(x) = 16(x^{2} - 6x - 7)[/tex]

[tex]0 = 16(x^{2} - 6x - 7)[/tex]

[tex]0 = x^{2} - 6x - 7[/tex]

[tex]0 = (x - 7)(x + 1)[/tex]

[tex]x = -1, 7[/tex]

Because time can only be positive, the answer is 7 seconds.

Answer:

7 seconds.

Step-by-step explanation:

height h = 16(x^2-6x-7) = 0

x^2 - 6x - 7 = 0

(x - 7)(x + 1) = 0

x = 7 seconds (we ignore the negative).

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