 Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 30.0 m/s to each pellet. Gun A is fired straight upwards, with the pellet going up and then falling back down, eventually hitting the ground beneath the cliff. Gun B is fired straight downwards. In the absence of air resistance, how long after pellet B hits the ground does pellet A hit the ground?★

To solve this problem, let's break it down step by step.

### Given:
- Initial speed of both pellets (\(v_0\)) = 30.0 m/s
- The height of the cliff is \(h\), which we will use as a variable.
- Acceleration due to gravity (\(g\)) = 9.8 m/s²
- Gun A is fired upwards, and Gun B is fired downwards.

### Step 1: Time for Pellet B to Hit the Ground
Pellet B is fired directly downward, so its initial velocity is in the same direction as gravity. The time \(t_B\) it takes to reach the ground can be calculated using the following kinematic equation:

\[
h = v_0 t_B + \frac{1}{2} g t_B^2
\]

Since \(h\) is the height and \(v_0\) is directed downward:

\[
h = 30.0 \, \text{m/s} \times t_B + \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times t_B^2
\]

### Step 2: Time for Pellet A to Hit the Ground
Pellet A is fired upward, so it will rise to a certain height before falling back down. The time can be split into two parts:
1. Time \(t_1\) to reach the highest point (where the velocity becomes zero).
2. Time \(t_2\) to fall back down from the highest point and hit the ground.

#### 2.1: Time to Reach the Maximum Height ( \(t_1\) )
At the maximum height, the final velocity becomes zero. Using the equation:

\[
v_f = v_0 - g t_1
\]

Setting \(v_f = 0\):

\[
0 = 30.0 \, \text{m/s} - 9.8 \, \text{m/s}^2 \times t_1
\]

Solving for \(t_1\):

\[
t_1 = \frac{30.0 \, \text{m/s}}{9.8 \, \text{m/s}^2} \approx 3.06 \, \text{seconds}
\]

#### 2.2: Time to Fall Back to the Ground ( \(t_2\) )
The pellet now falls from the maximum height, where the height \(h'\) is:

\[
h' = v_0 \times t_1 - \frac{1}{2} g t_1^2
\]

Plugging in the values:

\[
h' = 30.0 \, \text{m/s} \times 3.06 \, \text{seconds} - \frac{1}{2} \times 9.8 \, \text{m/s}^2 \times (3.06 \, \text{seconds})^2
\]

Calculate \(h'\):

\[
h' \approx 45.9 \, \text{m}
\]

The total height the pellet falls back down is \(h + h'\). The time \(t_2\) it takes to fall this distance is:

\[
h + h' = \frac{1}{2} g t_2^2
\]

The total time for Pellet A to hit the ground is:

\[
t_A = t_1 + t_2
\]

### Step 3: Calculate the Time Difference
The time difference between when Pellet A and Pellet B hit the ground is:

\[
\Delta t = t_A - t_B
\]

Using the earlier equations, calculate this difference.

### Simplified Approach

Since Pellet A goes up and comes back down with the same initial velocity but in opposite directions, the total time for A to reach the ground can be broken down as:

- The time taken to reach the original point from which it was fired upwards is the same as the time taken for B to hit the ground.
- The extra time is the time Pellet A takes to fall from the original point to the ground.

So, the difference in time is just \(t_1\), the time it takes for A to reach the highest point:

\[
\Delta t \approx 3.06 \, \text{seconds}
\]

Thus, **Pellet A hits the ground approximately 3.06 seconds after Pellet B**.

Would you like more details or have any questions? Here are some related questions for further practice:

1. What would happen to the time difference if the cliff height is doubled?
2. How would air resistance affect the time difference between when Pellet A and Pellet B hit the ground?
3. If Gun A is fired at an angle instead of straight up, how would this change the time difference?
4. Calculate the height of the cliff if Pellet B hits the ground after 4 seconds.
5. If both guns were fired horizontally, how would the time to hit the ground change?

**Tip:** When dealing with projectile motion, breaking down the motion into vertical and horizontal components simplifies the problem.

Explore a physics problem involving projectile motion: determine the time difference between when two pellets, fired upwards and downwards with identical initial speeds, hit the ground. Includes step-by-step calculations and formulas for projectile motion. Suitable for high school students.

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