Final answer:
To find how long the camera takes to reach the ground and its velocity just before it lands, we use the equations of motion for free falling objects with an initial velocity of -2.1 m/s and solve for time and final velocity.
Step-by-step explanation:
Time and Velocity of a Falling Camera
To solve both parts of this student's question, we need to use the equations of motion for an object in free fall under gravity. Since the hot-air balloon is descending at a rate of 2.1 m/s when the camera is dropped, we will consider the initial velocity for the camera of -2.1 m/s (negative because it is descending).
Part A: Time to Reach the Ground
We can use the second equation of motion, which relates initial velocity (u), acceleration due to gravity (g), distance (s), and time (t):
- s = ut + ½gt²
- 42 m = -2.1 m/s * t + ½(9.8 m/s²)t²
This is a quadratic equation in terms of time (t). Solving for t gives us the time it takes for the camera to hit the ground.
Part B: Velocity Just Before Landing
The final velocity (v) of the camera just before it hits the ground can be found using the first equation of motion:
- v = u + gt
- The negative initial velocity and the positive value of g (direction of the gravity) will determine the final value of the velocity. Since the initial velocity is in the opposite direction of gravity, the value of g assists to slow down the camera's descending speed until it reaches zero, after which it accelerates towards the ground.
Both the time to reach the ground and the final velocity right before the camera lands will be calculated using the above equations.