Rocket Launched Up When a Stone is Dropped Down

[putongren]

Homework Statement

At the base of a vertical cliff, a model rocket, starting from rest, is launched upwards at t = 0 with a time-varying acceleration given by

a(t) = A - Bt

where A and B are positive constants. Also at t = 0, a small stone is released form rest from the top of the cliff at a height h directly above the rocket. (This height h is higher than the maximum height reached by the rocket.). The stone hits the rocket at the instant when the rocket reaches its maximum height. The gravitational acceleration of magnitude g, is down dropped in terms of constants A, B and g.

Relevant Equations

Total Distance = Distance traveled by stone + Distance traveled by rocket

Total Distance = distance traveled by stone + distance traveled by rocket.

Total Distance = h
Distance traveled by stone = x_stone = (1/2)gt²


Distance traveled by rocket:
a_rocket(t) = A - Bt
v_rocket(t) = At - (1/2)Bt²
x_rocket(t) = (1/2)At² - (1/6)Bt³

In order to find the maximum height of the rocket, let v(t) = 0

So, 0 = At - (1/2)Bt²
0 = t(A - (1/2)Bt)
Therefore, t = (0, 2A/B)
The maximum of the rocket occurs when t = 2A/B.

Plug t = 2A / B into the Total Distance eqn:

h = (1/2)g(2A/B)² + (1/2)A(2A/B)² - (1/6)B(2A/B)³

 

[kuruman]

You described a physical situation. Fine. Is there a specific question that you want to answer given this situation? If so, what is it?

Obviously, the quantity h that you found in the last line is not the same as h in the drawing.

 

[putongren]

Can you critique and offer suggestions on how I can get the correct value for h? I already offered my personal solution for it. Since you say it's wrong, then I need help in obtaining the correct solution.

 

[kuruman]

The h that you calculated is the position of the rocket at the time it reaches maximum height. You stated yourself that "(This height h is higher than the maximum height reached by the rocket.)" Which h are you looking for, maximum height of the rocket or the initial height from which the ball is dropped?

You need to say with an equation that, at the time when the rocket reaches maximum height above ground, the stone is at that same height above ground.

 

[haruspex]

The h that you calculated is the position of the rocket at the time it reaches maximum height

Looks to me that @putongren has been consistent in using h for the cliff height.

 

[pbuk]

h = (1/2)g(2A/B)² + (1/2)A(2A/B)² - (1/6)B(2A/B)³

Yes that is correct. Were you required to find h to answer the question?

The gravitational acceleration of magnitude g, is down dropped in terms of constants A, B and g.

This sentence does not make sense. Perhaps there is a problem with translation?

 

[kuruman]

Looks to me that @putongren has been consistent in using h for the cliff height.

After looking at this with a clearer mind in the morning, I agree.

 

[putongren]

Yes that is correct. Were you required to find h to answer the question?



This sentence does not make sense. Perhaps there is a problem with translation?

Sorry I was typing to question directly from the problem set and I typed the sentence wrong. I will retype that specific sentence when I get home.

 

[putongren]

OK, back from work. This is the sentence I meant to type:

The gravitational acceleration of magnitude g is downward. You may neglect air resistance. Determine an expression for the initial height h from which the stone was dropped in terms of the constants A, B and g.