Projectile Launched from Moon: Altitude for 3/4 Speed

[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?

can someone please help me with this! i keep getting 1.96x10^5 m but it says that is not the correct answer

thanks

 

[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)?

 

[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?

 

[maverick_starstrider]

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

 

[HallsofIvy]

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"?

 

[ibcnunabit]

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"?

1.63 m/s^2

 

[ibcnunabit]

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"?

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.

 

[turin]

(assuming you're allowed to use mgh)

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.

 

[ralilu]

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)?

once i learned 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...