What went wrong in calculating the mass of planet Mongo?

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The discussion revolves around calculating the mass of planet Mongo based on measurements of a stone thrown upward. The initial calculations involved determining the acceleration of the stone using the formula V=V_o + at, but confusion arose regarding the final velocity when the stone returns to the ground. Participants pointed out that the final velocity at maximum height should be zero, which led to a reevaluation of the acceleration due to gravity. The original poster realized their calculated gravity was less than 4 m/s², indicating an error in their approach. Ultimately, clarifications on the correct use of initial and final velocities helped resolve the misunderstanding.
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Q:
Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: a 2.50-kg stone thrown upward from the ground at 15.0 m/s returns to the ground in 7.00 s; the circumference of Mongo at the equator is 1×10^5 km; and there is no appreciable atmosphere on Mongo. What is the mass of Mongo?

well i started out by finding the acceleration of the rock. I used V=V_o + at and from the problem, v_o = 15, and t=7.

Then I used F=ma to get the gravity force.

Then I used the general formula for force due to gravity. from the problem I used: R=1*10^5 / 2*pi (because they gave us the circumference) everything else is pretty self-explanitory. Since I have F, G, the mass of the rock, and R, I solved for M and got the wrong answer. Could anyone tell me where I went wrong?
 
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I used V=V_o + at and from the problem, v_o = 15, and t=7.
How do you know v?
 
It would help a little if you put some numbers in. A shot in the dark:
If you got an acceleration of gravity (a) that is less than 4\frac{m}{s^2} then you calculated that incorrectly.
 
thrown upward from the ground at 15.0 m/s returns to the ground in 7.00 s
Is this not an acceptable value for v_o? and, if it returned to the ground, wouldn't v be 0?
 
Is this not an acceptable value for v_o? and, if it returned to the ground, wouldn't v be 0?

No, v would take that value for maximum height.
 
...yeah I got a value less than 4...
 
So to get the equation to work out, I would just halve the time right? That would give me a V=0 at the max height. and an A of about 4.3m/s
 
Thanks a lot for the help guys, I got it now.
 
Berislav said:
How do you know v?
V = -V_o (?)
 
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V = -V_o (?)
The OP didn't state that, which lead me to think that he overlooked it.
 
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