Question

Invincible Cases manufactures cases that protect electronics against accidental drops. While testing a new model, engineers place a phone into the case and drop it onto a concrete floor from a height of 1.5 meters.
Which equation can you use to find how many seconds the phone falls before hitting the ground?
If an object falls from s meters above the ground, the object's height in meters, h, after t seconds can be modeled by the formula h=
–
4.9t2+s.
To the nearest tenth of a second, how long does the phone fall before hitting the ground?
seconds

Answer 1

To find the time \( t \) it takes for the phone to hit the ground, we can use the formula given for height \( h \):

\[
h = -4.9t^2 + s
\]

In this case, the height \( s \) from which the phone is dropped is 1.5 meters. When the phone hits the ground, the height \( h \) will be 0 meters. Therefore, we can set the equation to 0 and solve for \( t \):

\[
0 = -4.9t^2 + 1.5
\]

Rearranging the equation gives us:

\[
4.9t^2 = 1.5
\]

Next, divide both sides by 4.9:

\[
t^2 = \frac{1.5}{4.9}
\]

Now, calculate \(\frac{1.5}{4.9}\):

\[
t^2 \approx 0.30612244898
\]

Now take the square root of both sides to solve for \( t \):

\[
t \approx \sqrt{0.3061} \approx 0.553
\]

To find \( t \) to the nearest tenth of a second:

\[
t \approx 0.6
\]

Thus, the phone falls for approximately **0.6 seconds** before hitting the ground.