Volcano in mars ejects rocks. Straight line motion

AI Thread Summary
A volcano on Mars can eject rocks to a height that can be calculated using the same initial velocity as a volcano on Earth, but with Mars' lower gravity of 3.71 m/s² compared to Earth's 9.8 m/s². The maximum height on Earth is represented as H, while the height on Mars can be expressed as 2.64H. The calculations involve using the equations of motion, where the height is derived from the initial velocity and gravitational acceleration. The ratio of heights simplifies to the inverse ratio of the gravitational accelerations. Understanding when to take this ratio is crucial for solving similar problems.
Toranc3
Messages
189
Reaction score
0

Homework Statement



A certain volcano on Earth can eject rocks vertically to a maximum height .

Part A: How high (in terms of H ) would these rocks go if a volcano on Mars ejected them with the same initial velocity? The acceleration due to gravity on Mars is 3.71 m/s^2 , and you can neglect air resistance on both planets.


Homework Equations



y=yo + vo*t + 1/2 *g*t^(2)

vy=vo + g*t

The Attempt at a Solution



y=yo + vo*t - 1/2 *g*t^(2)


vy=vo + g*t
0=vo-g(mars)*t

t=vo/g
substitute this in y=yo + vo*t - 1/2 *g*t^(2)

to get y=1/2 * vo^(2)/g

After this I am stuck. The answer is 2.64H or 2.64Y


 
Physics news on Phys.org
Toranc3 said:

Homework Statement



A certain volcano on Earth can eject rocks vertically to a maximum height .

Part A: How high (in terms of H ) would these rocks go if a volcano on Mars ejected them with the same initial velocity? The acceleration due to gravity on Mars is 3.71 m/s^2 , and you can neglect air resistance on both planets.


Homework Equations



y=yo + vo*t + 1/2 *g*t^(2)

vy=vo + g*t

The Attempt at a Solution



y=yo + vo*t - 1/2 *g*t^(2)


vy=vo + g*t
0=vo-g(mars)*t

t=vo/g
substitute this in y=yo + vo*t - 1/2 *g*t^(2)

to get y=1/2 * vo^(2)/g

After this I am stuck. The answer is 2.64H or 2.64Y

g on Earth is 9.8m/s^2. What's the ratio of y=1/2 * vo^(2)/g between Mars and Earth?
 
Dick said:
g on Earth is 9.8m/s^2. What's the ratio of y=1/2 * vo^(2)/g between Mars and Earth?

Ah I see now thanks. However how do you know when to take the ratio? I did not even think about that.
 
Toranc3 said:
Ah I see now thanks. However how do you know when to take the ratio? I did not even think about that.

The height on Earth is H=(1/2)v0^2/(9.8m/s^2). The height on Mars will be (1/2)v0^2/(3.71m/s^2). If you divide the two then everything will cancel except for the ratio of the two values of g.
 
Dick said:
The height on Earth is H=(1/2)v0^2/(9.8m/s^2). The height on Mars will be (1/2)v0^2/(3.71m/s^2). If you divide the two then everything will cancel except for the ratio of the two values of g.

Got it and thank you.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top