# Projectile on the moon

## Homework Statement

A projectile launched vertically from the surface of the Moon rises to an altitude of 370 km. What was the projectiles initial speed?

## Homework Equations

Vf ^2 = Vi^2 + 2ad

## The Attempt at a Solution

distance = 370km x 1000 = 3.7E 5 meters
solving for Vi = sqroot of Vf^2 - 2ad
sqrt of 0 - -1.63 x 3.7E5 = 777m/s

ideasrule
Homework Helper
Check your calculations. 1.63 m/s^2 is equal to a, not 2a.

alright, i tried this vi=sqrt(2(1.63)(3.7)=1098, but thats wrong too, i still cant find out what im doing wrong

ideasrule
Homework Helper
1098 m/s is correct, unless the problem wanted you to take into consideration differences in acceleration due to the object's increasing altitude.

i dont know, my hw is online, so i just type in an answer, and it ells me whether im right or wrong, and it said i was wrong. So how would i compensate for the difference?

ideasrule
Homework Helper
You use the conservation of energy: initial kinetic energy + initial potential energy = final kinetic energy + final potential energy. Make sure to use U=-GMm/r for potential energy, not U=mgh.

what masses should i use?

ideasrule
Homework Helper
Well, in U=-GMm/r, M and m are the two attracting masses. Here, one of them is the Moon and the other is the projectile. You'll find that the mass of the projectile cancels out, so you don't need to know what it is.

ideasrule
Homework Helper
"r" is the distance from the center of the planet/moon to the projectile. So yes, you add the radius of the moon.

i got 1.01 m/s, which is wrong
0 + G(6.67E-11*7.36E22)/21074000^2 = 1/2vi^2 + G(6.67E-11*7.36E22)/1737400^2

It's U=-GMm/r not U=-GMm/r^2 f

just for easier ways to put it Kf=a Uf=b Ki=c Ui=d
it should look like this
c= -b + d