Question

Abdoulaye launches a toy rocket from a platform. The height of the rocket in feet is given by h, equals, minus, 16, t, squared, plus, 16, t, plus, 60h=−16t
2
+16t+60 where tt represents the time in seconds after launch. How long is the rocket in the air?

Answer 1

To determine how long the rocket is in the air, we need to find the time \( t \) when the height \( h \) becomes 0 (when the rocket hits the ground). The height of the rocket is given by the equation:

\[
h = -16t^2 + 16t + 60
\]

To find the time when the rocket is on the ground, we set \( h = 0 \):

\[
0 = -16t^2 + 16t + 60
\]

We can rearrange this equation:

\[
-16t^2 + 16t + 60 = 0
\]

Dividing the entire equation by -4 to simplify:

\[
4t^2 - 4t - 15 = 0
\]

Now, we can use the quadratic formula, \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). In our equation, \( a = 4 \), \( b = -4 \), and \( c = -15 \).

Calculating the discriminant:

\[
b^2 - 4ac = (-4)^2 - 4(4)(-15) = 16 + 240 = 256
\]

Now applying the quadratic formula:

\[
t = \frac{-(-4) \pm \sqrt{256}}{2(4)} = \frac{4 \pm 16}{8}
\]

This gives us two potential solutions for \( t \):

1. \( t = \frac{4 + 16}{8} = \frac{20}{8} = 2.5 \) seconds
2. \( t = \frac{4 - 16}{8} = \frac{-12}{8} = -1.5 \) seconds (not physically meaningful since time cannot be negative)

Thus, the time the rocket is in the air is:

\[
\boxed{2.5} \text{ seconds}
\]