Question

A model rocket 50 meters above group is falling at a constant rate of 4 meters per second while another model rocket is 20 meters above group is rising at a constant rate of 8 meters per second. After how many seconds will the 2 model rockets be the same height above the ground?

Answer 1

Let's denote the height of the first model rocket (falling) as \( h_1 \) and the height of the second model rocket (rising) as \( h_2 \).

The first rocket starts at a height of 50 meters and is falling at a rate of 4 meters per second. After \( t \) seconds, its height can be represented as:

\[
h_1 = 50 - 4t
\]

The second rocket starts at a height of 20 meters and is rising at a rate of 8 meters per second. After \( t \) seconds, its height can be represented as:

\[
h_2 = 20 + 8t
\]

We want to find the time \( t \) when both rockets are at the same height, so we set \( h_1 \) equal to \( h_2 \):

\[
50 - 4t = 20 + 8t
\]

Now, we will solve for \( t \):

1. Start by adding \( 4t \) to both sides:

\[
50 = 20 + 12t
\]

2. Next, subtract 20 from both sides:

\[
30 = 12t
\]

3. Finally, divide by 12:

\[
t = \frac{30}{12} = 2.5
\]

Thus, the two model rockets will be at the same height above the ground after **2.5 seconds**.

To confirm, we can calculate the heights at \( t = 2.5 \) seconds:

- For the first rocket:

\[
h_1 = 50 - 4(2.5) = 50 - 10 = 40 \text{ meters}
\]

- For the second rocket:

\[
h_2 = 20 + 8(2.5) = 20 + 20 = 40 \text{ meters}
\]

Both rockets are at a height of 40 meters after 2.5 seconds, verifying our solution. The answer is:

**2.5 seconds**.