# Speed in Space

1. Jul 21, 2009

### tgvaughn

A projectile is launched vertically from the surface of the Moon with an initial speed of 1210 m/s. At what altitude is the projectile's speed three-fourths its initial value?

thanks

2. Jul 21, 2009

### maverick_starstrider

This should really be in the hwk section. However, yes, your answer is incorrect. Think of it this way:

$\frac{1}{2}mv_1^2=\frac{1}{2}m (\frac{3v_1}{4})^2+mgh$ (assuming you're allowed to use mgh). I assume you've covered conservation of energy (or are you still only doing kinematics)?

3. Jul 21, 2009

### tgvaughn

yes we have covered conservation of energy. however when i solve using this equation i get the answer 195885, which is the same thing as 1.96x10^5 right?

4. Jul 21, 2009

### maverick_starstrider

Well you're making an algebraic mistake somewhere. I'm getting 3.3x10^4

5. Jul 22, 2009

### HallsofIvy

Staff Emeritus
And, of course, since we are talking about the moon and not the earth, you have to use the value of "g" that is correct for the moon, not 9.81 m/s2.

What is that "g"?

6. Jul 22, 2009

### ibcnunabit

1.63 m/s^2

7. Jul 22, 2009

### ibcnunabit

More precisely, an unconfirmed source says 1.62631, but I haven't done the calculation or found corroboration--but it's close enough to the approximation I do know that I don't doubt it.

8. Jul 22, 2009

### turin

You don't need to make an assumption; you can determine this by comparing to the more general expression. However, even if mgh turns out to be a good approximation, you might as well use the more general expression to begin with: U=GMm/r^2.

9. Jul 22, 2009

### ralilu

once i learnt the conservation of energy...physics became a whole lot easier coz i could basicly solve almost all problems using a few basic energy equations...instead of kinematics...