A rock is tossed straight up from the ground with a speed of 10 m/s. When it returns, it falls into a hole 10 m deep. How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

To solve this problem, we can break it into two parts:

1. The time the rock takes to reach its highest point and return to the ground level.
2. The time the rock takes to fall from ground level into the hole.

### Step 1: Motion up and down (before reaching ground level)

The rock is tossed up with an initial velocity of \( v_0 = 10 \, \text{m/s} \). We can use the following kinematic equation to determine the time for the rock to reach its highest point:

\[
v = v_0 - g t
\]

At the highest point, the velocity \( v = 0 \), and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \)).

Solving for time \( t_1 \) (the time to reach the highest point):

\[
0 = 10 - 9.8 t_1
\]

\[
t_1 = \frac{10}{9.8} \approx 1.02 \, \text{seconds}
\]

Since the time to go up equals the time to come back down to ground level, the total time to return to the ground is:

\[
t_{\text{ground}} = 2 \times 1.02 = 2.04 \, \text{seconds}
\]

### Step 2: Falling into the hole

Now, the rock falls an additional \( 10 \, \text{m} \) into the hole. We can use the following equation to determine the time \( t_2 \) it takes to fall this distance:

\[
d = \frac{1}{2} g t_2^2
\]

Where \( d = 10 \, \text{m} \) and \( g = 9.8 \, \text{m/s}^2 \). Solving for \( t_2 \):

\[
10 = \frac{1}{2} \times 9.8 \times t_2^2
\]

\[
t_2^2 = \frac{20}{9.8} \approx 2.04
\]

\[
t_2 \approx \sqrt{2.04} \approx 1.43 \, \text{seconds}
\]

### Total time in the air

The total time the rock is in the air is the sum of the time to return to the ground and the time to fall into the hole:

\[
t_{\text{total}} = t_{\text{ground}} + t_2 = 2.04 + 1.43 = 3.47 \, \text{seconds}
\]

Thus, the rock is in the air for approximately **3.47 seconds**.

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Would you like more details on any part of the explanation?

Here are five related questions to consider:

1. What would happen if the initial speed of the rock was doubled?
2. How would the total time change if the hole was deeper?
3. What is the maximum height reached by the rock before it falls?
4. How would the time be affected if gravity was stronger (e.g., on another planet)?
5. Can you calculate the velocity of the rock as it hits the bottom of the hole?

**Tip:** When solving kinematics problems, dividing the motion into phases (e.g., upward, downward) helps manage the complexity!

This detailed solution explains how to calculate the total time a rock stays in the air after being tossed straight up at 10 m/s and falling into a 10-meter deep hole. The problem covers kinematics, free fall, and the use of key formulas in projectile motion. Ideal for students in Grades 9-12.

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