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.