## anonymous one year ago A certain volcano on earth can eject rocks vertically to a maximum height H. The acceleration due to gravity on Mars is 3.71 m/s2, and you can neglect air resistance on both planets. If the rocks are in the air for a time T on earth, for how long (in terms of T) will they be in the air on Mars?

1. Astrophysics

Write down all the variables which are present in this problem

2. Astrophysics

Think about kinematics for now, what does it mean maximum heigh h?

3. anonymous

So there is a previous part to this question which asks for the ratio between maximum height a rock will reach on Mars and that of a rock on Earth. $\frac{ Hm }{ H } = 2.64$in which Hm is the max height a rock on Mars will travel and H is that of a rock on Earth. I know that the initial velocity is 0 and I manipulated a kinematics equation to solve for that ratio. But what should I do to get time? Should I do the same steps incorporating 2.64 somewhere? I can tell you how I solved for 2.64.

4. Astrophysics

at maximum height vf = 0

5. Astrophysics

Yes, you can use the ratio to find the height on mars

6. anonymous

What is my firsrt step? Do I use the formula vf^2 = vi^2 + 2ad and substitute the ratio into d and also set vf = 0?

7. anonymous

I'm mostly lost on which formula is the most appropriate to use for this question.

8. Astrophysics

Was your first question in terms of h?

9. anonymous

Yes it was.

10. Astrophysics

So for that you took the ratio of the kinematic equation v^2 = v0^2+2ah then you should have 2.64H

11. anonymous

Yes that is what I did!

12. Astrophysics

Ok, so now use since you want time, what kinematic equation can we use?

13. anonymous

d = Vi(t) + (1/2)a(t)^2 since we know the acceleration on earth, Vi = 0 and we use 2.64 somehow for the displacement?

14. Astrophysics

Yes we use that equation, specifically $y-y_0=v_0t+1/2at^2$ the direction would be the y direction but I'm not to worries about putting subscripts, so the y-y0/ d = 0 so then we will be left with $0=v_0t+1/2at^2$

15. Astrophysics

The reason it is 0 is because it goes up and then it returns to the ground

16. anonymous

I see! Thanks. Now we need to find V0 to solve for t? Setting V0 to 0 only makes t = 0.

17. Astrophysics

You should end up with $(at)_E=(at)_M$ where and E and M are the planets

18. Astrophysics

You should note the answer is similar to a but the quantity of course is different

19. anonymous

So the answer was actually the same, but the method to get there was different. Thank you so much and thank you for being patient for me!

20. Astrophysics

Sounds good! And np :)