A rock is thrown in the air from the edge of a seaside cliff. Its height in feet is represented by f(x) = - 16(x ^ 2 - 8x - 9) where x is the number of seconds since the rock was thrown. The height of the rock is feet when it hits the water. How long does it take the rock to hit the water? A. 1 second B. 9 seconds C. 8 seconds D. 16 seconds
To find out how long it takes for the rock to hit the water, we need to find the value of x when f(x) = 0.
Given that f(x) = -16(x^2 - 8x - 9)
Setting f(x) = 0:
0 = -16(x^2 - 8x - 9)
0 = x^2 - 8x - 9
Now, we need to solve the quadratic equation x^2 - 8x - 9 = 0 for x.
Factoring the quadratic equation, we get:
0 = (x - 9)(x + 1)
Setting each factor to zero gives us the roots:
x - 9 = 0, x = 9
x + 1 = 0, x = -1
Since time cannot be negative, the rock must hit the water at x = 9 seconds.
Therefore, it takes 9 seconds for the rock to hit the water, which means the answer is B. 9 seconds.