Question

For his school's annual egg-drop contest, Nick created a cushioned egg container made of cotton balls and tin foil. In the first round, Nick dropped his egg from the school's second floor balcony 18 feet above the ground, and the egg survived!
Which equation can you use to find how many seconds it took for the egg to hit the ground?
If an object is dropped from s feet above the ground, the object's height in feet, h, t seconds after being dropped can be modeled by the formula h=
–
16t^2+s.

0=
–16t^2+18

18 –16t^2
To the nearest tenth of a second, how long did it take for the egg to hit the ground?
seconds

Answer 1

To determine how long it took for the egg to hit the ground when dropped from a height of 18 feet, you can use the formula for the height of the object:

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

where \( s \) is the initial height (in feet), \( h \) is the height after \( t \) seconds, and \( -16t^2 \) accounts for the effect of gravity.

In this scenario:
- \( s = 18 \)
- The egg hits the ground when \( h = 0 \).

So you set up the equation:

\[
0 = -16t^2 + 18
\]

Now, let's solve for \( t \):

1. Rearranging gives:
\[
16t^2 = 18
\]

2. Dividing both sides by 16:
\[
t^2 = \frac{18}{16} = \frac{9}{8}
\]

3. Taking the square root of both sides:
\[
t = \sqrt{\frac{9}{8}} = \frac{3}{\sqrt{8}} = \frac{3\sqrt{8}}{8}
\]

Calculating \( \sqrt{8} \) gives \( \sqrt{8} = 2\sqrt{2} \), so:

\[
t = \frac{3 \times 2\sqrt{2}}{8} = \frac{3\sqrt{2}}{4}
\]

Next, we can calculate \( t \) using a calculator:

1. Find \( \sqrt{2} \approx 1.414 \).
2. Thus, \( 3\sqrt{2} \approx 3 \times 1.414 = 4.242 \).
3. Then divide by 4:
\[
t \approx \frac{4.242}{4} \approx 1.061
\]

Rounding to the nearest tenth, \( t \approx 1.1 \) seconds.

So, the answer is:

**It took approximately 1.1 seconds for the egg to hit the ground.**