Velocity-Time Curve for Balloon-Dropped Stones | A/L Math Help

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The discussion revolves around a math problem involving a balloon accelerating upwards while dropping two stones. The first stone (P) is dropped at time t = 0 and lands after time T, while the second stone (Q) is dropped after time t and falls from a higher point. Participants clarify the conditions of the problem and the equations involved, particularly focusing on the velocity-time curves for both stones relative to the balloon. The expression "t(g+a)(T+t/2)" is debated, with one participant suggesting it cannot be proven as it lacks equation status. The conversation emphasizes understanding the relationship between the balloon's acceleration and the stones' motion.
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Hi,
I have my A/L s tomorrow and I need help in a maths question.
If a baloon is going strait upwards in an accelaration "a" and it drops a stone(P) and after the time "t" the baloon drops another stone(Q). The first stone lands on the Earth after droping the first stone at the time "T".Draw the velocity time curve for these two stones and it should be drawn relative to the baloon.
Prove that "t(g+a)(T+t/2)"
 
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campa said:
Hi,
I have my A/L s tomorrow and I need help in a maths question.
If a baloon is going strait upwards in an accelaration "a" and it drops a stone(P) and after the time "t" the baloon drops another stone(Q). The first stone lands on the Earth after droping the first stone at the time "T".Draw the velocity time curve for these two stones and it should be drawn relative to the baloon.
Prove that "t(g+a)(T+t/2)"

Hi campa. I'm not sure I understand the question.

The balloon is going upwards with acceleration a.

1) Suppose it drops P at some time x?
2) It drops Q at time x+t?
3) P lands at time x+T?

Are the above 3 correct?

Also "t(g+a)(T+t/2)" cannot be proven as it is not an equation, just an algebraic expression.
 


Hello! I'm happy to help with your A/L math question.

To start, let's define some variables:
- a = acceleration of the balloon (in m/s^2)
- t = time it takes for the balloon to drop the second stone (in seconds)
- T = time it takes for the first stone to land on Earth (in seconds)
- g = acceleration due to gravity (in m/s^2)

Now, let's look at the velocity-time curve for the two stones:

- The first stone (P) is dropped by the balloon at time t = 0. Its initial velocity is 0 m/s, since it was dropped from rest. As it falls towards the Earth, its velocity will increase due to the acceleration of gravity (g).

- The second stone (Q) is dropped by the balloon at time t = t. Its initial velocity is also 0 m/s, but it starts falling from a higher point than the first stone. Therefore, its velocity will also increase due to the acceleration of gravity (g).

- The balloon, which is still going upwards with an acceleration of a, will also have a velocity-time curve. However, since we are only interested in the stones, we can ignore this curve for now.

Now, let's look at the time it takes for the first stone to land on Earth (T). This time can be calculated using the formula: T = √(2h/g), where h is the initial height of the stone (in meters). Since the first stone was dropped at the same height as the balloon, we can say that h = 0. Plugging this into the formula, we get T = 0. This means that the first stone lands on Earth at the same time it was dropped from the balloon (t = T).

Next, let's look at the time it takes for the second stone to land on Earth. This time can be calculated using the formula: T + t = √(2h/g). Since the second stone was dropped at a higher height than the first stone, we can say that h = a(t/2)^2 (using the formula for displacement under constant acceleration). Plugging this into the formula, we get T + t = √(2a(t/2)^2 / g). Simplifying, we get T + t = √(t^2a/g). Squaring both sides, we get (T +
 
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