Question

An object is launched from the ground. The object’s height, in feet, can be described by the quadratic function h(t) = 80t – 16t2, where t is the time, in seconds, since the object was launched. When will the object hit the ground after it is launched? Explain how you found your answer.

Answer 1

To determine when the object hits the ground, we need to find the time \( t \) when the height \( h(t) \) equals 0. We start with the function:

\[
h(t) = 80t - 16t^2
\]

We set this equal to zero:

\[
80t - 16t^2 = 0
\]

Next, we can factor out the common term. Both terms on the left side have \( t \) as a factor:

\[
t(80 - 16t) = 0
\]

This gives us two potential solutions:

1. \( t = 0 \) (which represents the time of launch)
2. \( 80 - 16t = 0 \)

Now, we solve the second equation for \( t \):

\[
80 - 16t = 0
\]
\[
16t = 80
\]
\[
t = \frac{80}{16} = 5
\]

Thus, the object will hit the ground after \( 5 \) seconds.

In summary, we found the solution by setting the height function equal to zero and factoring the resulting equation, which allowed us to solve for the time. The object hits the ground at \( t = 5 \) seconds after it is launched.