Velocity-time graph for stones dropped from rising balloon

[campa]

Hi,
I have my A/L s tomorrow and I need help in a maths question.
If a balloon is going strait upwards in an accelaration "a" and it drops a stone(P) and after the time "t" the balloon 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 balloon.
Prove that "t(g+a)(T+t/2)"

 

[learningphysics]

Hi,
I have my A/L s tomorrow and I need help in a maths question.
If a balloon is going strait upwards in an accelaration "a" and it drops a stone(P) and after the time "t" the balloon 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 balloon.
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.

 

[thepatientmental]

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 +