What Initial Speed Launches a Projectile to 270 km on the Moon?

[wadini]

A projectile launched vertically from the surface of the Moon rises to an altitude of 270 km.
What was the projectile's initial speed in m/s?

okay so I keep getting this answer wrong but I am pretty sure I am doing it correctly...

this is what I am doing:

2*6.67*7.35*270/3.02 and then all of that multiplied by 10^2 and then that answer square rooted and I get 936.27 m/s ...but that is not correct...HELP! what am I missing??

 

[CompuChip]

Can you maybe explain where you got all those numbers from?
What is 6.67, what is 7.35, what is 3.02, where do the 2 and 10^2 come from and why square-root it?

i.e. post your formula and identify the variables. Now it's just a string of numbers without physical meaning.

 

[wadini]

Vi- square root 2GMmH/(1.74*10^6)2

7.35 *10^22 kg is the mass of the moon Mm
Radius of the moon Re=1.74*10^6
g= 6/67*10^-11
height reached by the projectile h= 260km = 270* 10^3
I plugged all of that into the equation above and got the answer previously stated but it is wrong.

 

[wadini]

sorry Vintial = **

 

[MattRSK]

Sounds like a kinematic equation problem to me.

 

[rl.bhat]

Apply the conservation of energy theorem. When the projectile reaches the maximum height, the KE of projectile is equal to the change in PE.
PE on the surface of the Earth = GMm/R. When it reaches the maximum height, the PE = GMm/(R+h)