Question

Paige launched a ball using a catapult she built. The height of the ball (in meters above the ground)
\[t\] seconds after launch is modeled by
\[h(t)=-5t^2+40t\]
Paige wants to know when the ball will hit the ground.
1) Rewrite the function in a different form (factored or vertex) where the answer appears as a number in the equation.
\[h(t)=\]
2) How many seconds after launch does the ball hit the ground?
seconds

Answer 1

To find out when the ball will hit the ground, we first need to rewrite the height function \(h(t)\) and then find the time at which the ball reaches a height of 0 meters.

1) The original function is given by:
\[
h(t) = -5t^2 + 40t
\]

To rewrite this function in a factored form, we can factor out the common terms:

\[
h(t) = -5(t^2 - 8t)
\]

Next, we can complete the square inside the parentheses. The expression \(t^2 - 8t\) can be rewritten by completing the square:

\[
t^2 - 8t = (t - 4)^2 - 16
\]

So, substituting this back into the equation gives us:
\[
h(t) = -5((t - 4)^2 - 16)
\]
\[
h(t) = -5(t - 4)^2 + 80
\]

Now we have the function in vertex form:
\[
h(t) = -5(t - 4)^2 + 80
\]

The height \(h(t)\) reaches ground level (0 meters) when:
\[
-5(t - 4)^2 + 80 = 0
\]

2) To find when the ball hits the ground, we can set \(h(t) = 0\) and solve for \(t\):

Starting from the factored vertex form:
\[
-5(t - 4)^2 + 80 = 0
\]
Adding \(5(t - 4)^2\) to both sides yields:
\[
5(t - 4)^2 = 80
\]
Dividing both sides by 5 gives:
\[
(t - 4)^2 = 16
\]
Taking the square root of both sides results in:
\[
t - 4 = 4 \quad \text{or} \quad t - 4 = -4
\]
Thus, solving for \(t\):
1. \(t - 4 = 4 \Rightarrow t = 8\)
2. \(t - 4 = -4 \Rightarrow t = 0\)

The ball is launched at \(t = 0\) and hits the ground at \(t = 8\) seconds.

Thus, the answers are:
1) Rewritten function:
\[
h(t) = -5(t - 4)^2 + 80
\]

2) The ball hits the ground:
\[
t = 8 \text{ seconds}
\]