Question

an object is thrown upward from the top of an 80-foot building with an initial velocity of 64 feet per second. the height h of the object after t seconds is given by the quadratic equation h = -16t² + 64t + 80. when will the object hit the ground? the object will hit the ground when the time is □ seconds.

Explanation:

Step1: Set h = 0 (ground level)

We need to solve the equation \(0 = -16t^2 + 64t + 80\). First, we can divide the entire equation by -16 to simplify it:
\(\frac{0}{-16}=\frac{-16t^2}{-16}+\frac{64t}{-16}+\frac{80}{-16}\)
Which simplifies to \(0 = t^2 - 4t - 5\)

Step2: Factor the quadratic equation

We factor \(t^2 - 4t - 5\). We need two numbers that multiply to -5 and add to -4. Those numbers are -5 and 1. So,
\(t^2 - 4t - 5=(t - 5)(t + 1)=0\)

Step3: Solve for t

Set each factor equal to zero:

• For \(t - 5 = 0\), we get \(t = 5\)
• For \(t + 1 = 0\), we get \(t = -1\)

Since time cannot be negative, we discard \(t = -1\).

Answer:

5