To determine the speed with which the camera hits the ground, we can break the problem into two parts:

1. Determine the time it takes for the camera to fall from Luke's hands to the ground.
2. Determine the speed with which the camera hits the ground.

Part 1: Time of Free Fall

We will use the following kinematic equation to calculate the time of free fall:

$d = v_i t + \frac{1}{2} a t^2$

Where:

• $d = 48.3 \, \text{m}$ is the height from which the camera was dropped,
• $v_i = 0 \, \text{m/s}$ is the initial velocity (since Luke fumbled it from rest),
• $a = 9.8 \, \text{m/s}^2$ is the acceleration due to gravity,
• $t$ is the time we want to find.

Since $v_i = 0$, the equation simplifies to:

$d = \frac{1}{2} a t^2$

Solving for $t$:

$t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2 \times 48.3}{9.8}} = \sqrt{9.857} \approx 3.14 \, \text{seconds}$

Part 2: Final Speed

Now that we know the time of free fall is approximately $3.14$ seconds, we can calculate the final speed of the camera when it hits the ground using the following kinematic equation:

$v_f = v_i + a t$

Where:

• $v_i = 0 \, \text{m/s}$ (initial velocity),
• $a = 9.8 \, \text{m/s}^2$,
• $t = 3.14 \, \text{seconds}$.

Thus,

$v_f = 0 + (9.8)(3.14) = 30.77 \, \text{m/s}$

Final Answer:

The camera hits the ground with a speed of approximately 30.77 m/s.

Would you like any details or further explanation?

Here are 5 related questions to expand on the concept:

1. How would the impact speed change if the camera were dropped from a higher altitude?
2. What is terminal velocity, and how does it differ from free fall speed?
3. How would air resistance affect the camera's speed as it falls?
4. What is the role of gravity in determining the final speed of falling objects?
5. Could the mass of the camera affect the falling speed if we considered air resistance?

Tip: When solving free fall problems, always remember to check if air resistance is considered, as it can significantly change the final speed.