Gravitational Potential Energy

[Biker]

vx=vo cos(theta)?

Yes now try to form an equation to get the maximum height this ship can get to using energy conservation

 

[Taniaz]

That's what I'm not sure of. I know the maximum height h = vo^2 sin ^ 2 (theta) / 2g is what we have for constant gravity
I know what vo is in terms of h but what happens to the sin^2(theta)/2g

h will only depend on vx so we know that vx=vo cos(theta) so vo=vx/cos(theta)

Plug it into h= vo^2 sin ^ 2 (theta) / 2g = ( vx /cos(theta))^2 sin^2(theta)
Not sure what happens with the 2g, do I just disregard it.
So h = vx sin^2 (theta) / cos^2 (theta) = vx tan^2 (theta)
Am I anywhere close? I don't know.

 

[Biker]

That's what I'm not sure of. I know the maximum height h = vo^2 sin ^ 2 (theta) / 2g is what we have for constant gravity
I know what vo is in terms of h but what happens to the sin^2(theta)/2g?

Not sure what you mean.. Clarify a bit more.

Anyway, You can't use this equation because it assumes that gravity is constant (Which we may approximate if we are not really far from a point). As Haruspex, we use energy conservation principle to have more accurate results.

So what you have to do is just ask yourself. If I am starting at the surface of the planet what kind of energy do I have? Assuming I am not going vertically, What kind of energy do I have at the maximum height?

So as you pointed out, we have velocity ##V_o## at the beginning and what is the potential energy?. Now at maximum height, you figured we only have horizontal velocity and what is the potential energy at that point?

From conservation of energy principle, we know that at any point the energy must be constant. Why not equate the two and form an equation?

 

[Biker]

You could just find the change in kinetic energy if it was going vertically and the change in kinetic energy if it wasn't and say that they don't equal each other or situation #1 > situation 2# which means it will not reach it. But it is a good practice to form equations. Physics is gained by practising.

 

[jbriggs444]

Lets remember what we studied about projectile motion. If the only force is gravity then the horizontal velocity vector is always constant.

Be careful. The horizontal velocity vector is not going to be constant when "horizontal" is changing. If we are considering a realistic scenario in which gravity drops off as a function of vertical position then we need to also consider that the surface of the Earth is a sphere and not a plane. "Horizontal" and "vertical" change as a function of horizontal position.

While the horizontal component of velocity does change over time, there is a different conserved quantity that does not. Can you name it?

 

[Biker]

Be careful. The horizontal velocity vector is not going to be constant when "horizontal" is changing. If we are considering a realistic scenario in which gravity drops off as a function of vertical position then we need to also consider that the surface of the Earth is a sphere and not a plane. "Horizontal" and "vertical" change as a function of horizontal position.

While the horizontal component of velocity does change over time, there is a different conserved quantity that does not. Can you name it?

Sorry for the last response, I blame high school physics. #Earthisflat (It is a good approximation if we are dealing with not very long horizontal distances but as they are astronauts and leaving the planet then yep as you said)

Not sure what you are suggesting at. I have thought of some ways but didn't reach any answer. Might want an insight on this :c

P.s Apologizing for the OP

 

[jbriggs444]

Not sure what you are suggesting at

Spoiler

Angular momentum

 

[Biker]

Spoiler

Angular momentum

Yep still haven't taken that. Thanks.

For the OP, You could use that to finish your answer. (If you couldn't figured it out.)