How Fast Does a Rock Fall on Jupiter?

[Coastal]

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

 

[ranger]

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:

v_f = v_i * at
d = v_i*t + 1/2*a*t²
v_i - initial velocity
d - distance

You first need to find the time it took to fall, then you can find the final velocity (v_f).

 

[Coastal]

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?

 

[ranger]

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 V_f is V_f/g. When you multiply g and t, you must obviously get V_f.

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.

 

[Coastal]

Yep, that worked fine. Thanks for your help!