Question

A ball is thrown in the air from a ledge. Its height in feet is represented by
f(x) =-16(x^2 - 5x - 6), where x is the number of seconds since the ball has been thrown. The height of the ball is O feet when it hits the ground.
How many seconds does it take the ball to reach the ground?
• 5 seconds
• 6 seconds
O 16 seconds
O 1 second

Answers

GPT-4o mini
To find out how many seconds it takes for the ball to reach the ground, we need to determine when its height \( f(x) = 0 \). We start with the equation:

\[
f(x) = -16(x^2 - 5x - 6)
\]

Setting \( f(x) \) to 0 gives us:

\[
-16(x^2 - 5x - 6) = 0
\]

Since \(-16\) is simply a constant multiplier, we can ignore it and set the quadratic part to zero:

\[
x^2 - 5x - 6 = 0
\]

Next, we can factor the quadratic equation. We are looking for two numbers that multiply to \(-6\) (the constant term) and add to \(-5\) (the coefficient of \(x\)). The numbers that satisfy this are \(-6\) and \(1\). Thus, we can factor the equation as follows:

\[
(x - 6)(x + 1) = 0
\]

Setting each factor equal to zero gives us the solutions:

1. \( x - 6 = 0 \) → \( x = 6 \)
2. \( x + 1 = 0 \) → \( x = -1 \)

Since time cannot be negative, we discard \( x = -1 \). Hence, the time it takes for the ball to reach the ground is:

\[
\boxed{6 \text{ seconds}}
\]