Very simple problem, cant get it

  • Thread starter Coastal
  • Start date
In summary, an astronaut on Jupiter drops a rock and calculates its speed just before it lands. His equation is wrong, but he gets the same answer as another equation that uses time and distance.
  • #1
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Homework Statement


An astronaut on Jupiter drops a rock straight downward from a height of 1.05 m. If the acceleration of gravity on Jupiter is 24.8 m/s2, what is the speed of the rock just before it lands?

Homework Equations


g=24.8m/s2
speed=distance/time

The Attempt at a Solution


At first i attempted to simply multiply 24.8 and 1.05, but realized that was probably wrong. I then calculated t by dividing 1.05 by 24.8 and got t=.042. Then used speed = distance/time to get s=1.05/.042 to get 1.04m/s. Does this sound correct, or is it way off? Would just like to know if this is the right method or if my answer is right. Thx
 
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  • #2
Your approach is wrong. If we divide (1.05m)/(24.8m/s2) we do not get the unit of second, instead we get second2, which isn't right for time. Also I'm not sure why you even want to divide distance by acceleration. You need to use a pair of kinematic equations for this:

vf = vi * at
d = vi*t + 1/2*a*t2
vi - initial velocity
d - distance

You first need to find the time it took to fall, then you can find the final velocity (vf).
 
Last edited:
  • #3
The problem I am working on corresponds with a problem in a book I have that also has an answer. I used to formula V= Square root of 2*g*x (formula of velocity as a function of position). This V was listed as the answer for the final velocity in the book. Instead, I manipulated the velocity as a function of time formula (v=gt) to t=v/g, to get the time (I got .29s), then put that into the formula Vf=Vi*at. I copied this method for the book problem and got a different answer than the one the book got.

So... is my formula not correct, or would Vf simply be the square root of 2gx?
 
  • #4
The problem I am working on corresponds with a problem in a book I have that also has an answer. I used to formula V= Square root of 2*g*x (formula of velocity as a function of position). This V was listed as the answer for the final velocity in the book.
Ah, yes. This was an easier formula to use rather than find time, then find the final velocity.
I manipulated the velocity as a function of time formula (v=gt) to t=v/g, to get the time (I got .29s), then put that into the formula Vf=Vi*at. I copied this method for the book problem and got a different answer than the one the book got.
Not sure exactly what you did, but the time it took for the object to reach Vf is Vf/g. When you multiply g and t, you must obviously get Vf.
So... is my formula not correct, or would Vf simply be the square root of 2gx?
Yup. You'd get the same answer if you used the two equations I gave you, or use square root of 2*a*d.
 
  • #5
Yep, that worked fine. Thanks for your help!
 

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