A hot-air balloon is rising vertically at a speed of 10 m s–1. An object is released from the balloon. The graph shows how the velocity of the object varies with time from when it leaves the balloon to when it reaches the ground four seconds later. It is assumed that the air resistance is negligible. (Diagram II)

Hence determine the object’s final displacement from its point of release from the balloon.
(c) Using the axes below, sketch a graph showing how the acceleration of the object changes during the time from when it leaves the balloon to when it hits the ground.
Mark any significant values on the axes. (Diagram III)
(a) Use the graph to:
(i) Show that the object continues to rise for a further 5 m after it is released.

(ii) Determine the total distance travelled by the object from when it is released from the balloon to when it reaches the ground.

Well, let's dive into some physics with a touch of humor, shall we?

(a)
(i) According to the graph, after 4 seconds, the velocity of the object becomes zero. However, we can see that the velocity-time graph does not immediately drop all the way down to the x-axis. So, if we follow the line horizontally from the point where velocity is zero, until we intersect the distance axis, we will find that the object needs to travel an additional 5 meters upwards before it stops.

(ii) To determine the total distance traveled by the object, we need to consider both the upward and downward distances. The upward distance can be calculated by finding the area under the velocity-time graph until the object reaches its maximum height. The downward distance can be calculated by finding the area under the velocity-time graph from the maximum height until the object hits the ground. By summing up these two distances, we will get the total distance traveled by the object.

Now, let's move on to sketching the graph of acceleration.

(c)
Now, I don't have the diagram in front of me, but I'll do my best to give you a mental picture. Imagine the vertical axis representing acceleration and the horizontal axis representing time. Given that the object is under the influence of gravity and there is no air resistance, we know that its acceleration will be a constant value, which happens to be approximately 9.8 m/s². So, throughout the entire journey from the point of release to hitting the ground, the acceleration will remain at a constant value of 9.8 m/s².

So, on our graph, we can have a straight line with a slope of 9.8, representing constant acceleration from the initial time to the final time.

I hope that helps, and remember, stay grounded even when you're floating in a hot-air balloon!

(a) (i) To show that the object continues to rise for a further 5 m after it is released, we need to find the displacement of the object after it is released from the balloon.

Looking at the graph, we can see that the velocity of the object is positive (above the time axis) for 1 second after it is released. Since velocity is the rate of change of displacement, the area under the velocity-time graph for this period will give us the displacement during this time.

The area under the graph is a triangle with base 1 second and height 10 m/s. The formula to find the area of a triangle is (1/2) * base * height.

Therefore, the displacement during this 1-second period is (1/2) * 1 s * 10 m/s = 5 m.

Since the object continues to rise for a further 5 m after it is released, we conclude that the object continues to rise for a further 5 m after it is released.

(a) (ii) To determine the total distance traveled by the object from when it is released from the balloon to when it reaches the ground, we need to find the total area under the velocity-time graph.

Breaking the graph into separate sections:

1. The area for the first 1 second (which we just calculated) is 5 m.
2. The next 2 seconds, the velocity becomes negative (below the time axis), indicating that the object is falling. The area for this part of the graph is a rectangle with a base of 2 seconds and a height of 10 m/s. The formula to find the area of a rectangle is base * height, so the area during this period is 2 s * 10 m/s = 20 m.
3. Finally, the object reaches the ground, indicated by the horizontal line on the graph. The area for this part is a rectangle with a base of 1 second and a height of 0 m/s. The area during this period is 1 s * 0 m/s = 0 m.

Adding up the areas, the total distance traveled by the object is 5 m + 20 m + 0 m = 25 m.

To determine the object's final displacement from its point of release from the balloon, we need to look at the graph provided. The graph shows the velocity of the object as a function of time.

(c) To sketch a graph showing how the acceleration of the object changes during the time from when it leaves the balloon to when it hits the ground, we need to analyze the velocity-time graph.

(a) (i) To show that the object continues to rise for a further 5 m after it is released, we need to find the time when the velocity becomes zero (when the object reaches its maximum height). From the graph, we can see that the object takes 2 seconds to reach its maximum height. At that point, its velocity is zero. We know that the object continues to rise for a further 5 m after being released because the graph shows a positive slope before the velocity becomes zero. This positive slope indicates that the object is still moving upwards. Therefore, we can conclude that the object continues to rise for a further 5 m after it is released.

(ii) To determine the total distance traveled by the object from when it is released from the balloon to when it reaches the ground, we need to find the area under the graph. From the graph, we can see that it takes a total of 6 seconds for the object to reach the ground.

To calculate the total distance traveled, we divide the graph into two parts: the first part is the upward journey (from the release to the maximum height) and the second part is the downward journey (from the maximum height to the ground).

1. Upward journey:
The area under the graph during the upward journey can be calculated as the area of a trapezoid (since velocity is changing linearly). The formula for the area of a trapezoid is A = (a + b) * h / 2, where a and b are the lengths of the parallel sides, and h is the height (time interval) between them.

For the upward journey, the parallel sides are 0m/s (at the release) and 10m/s (at the maximum height), and the time interval is 2 seconds.

So, the area of the trapezoid is A = (0 + 10) * 2 / 2 = 10 m.

2. Downward journey:
The downward journey can be represented as a rectangle with a base of 4 seconds (time interval) and a height of 10 m/s (the velocity remains constant during the downward journey).

So, the area of the rectangle is A = 10 * 4 = 40 m.

Therefore, the total distance traveled by the object is 10 m (upward journey) + 40 m (downward journey) = 50 m.

Hence, the total distance traveled by the object from when it is released from the balloon to when it reaches the ground is 50 m.

The answer is really complicated, maybe 0 but Im not sure.