Quote (ass666 @ Feb 8 2010 06:13am)

break the 27m/s into x and y components using sin and cos or whatever is easier for you.

Use the y component and divide it by the acceleration on the moon (9.8/6). That is its up time. Multiply that by two and its its total time for the trajectory until it hits the floor assuming curvature of the moon is negligible.

Get the x component and multiply it by the total air time and there you go =) distance.

Do the same thing but instead of using 9.8/6 use just 9.8 find the distance and take the difference.

hmm weird this is what i did..

earth

0= vyt+0.5gt^2

gt^2=2vyt

t=2vy/g 2(11.62)/9.8 = 2.37 s

x = (vx+vx)t/2

(24.92+24.92)(2.37)/2 = 59.06m

moon

0= vyt+0.5gt^2

gt^2=2vyt

t=2(vy)/(g/6) 2(11.62)/(9.8/6) = 14.228 s

x = (vx+vx)t/2

(24.92+24.92)(14.22)/2 = 354.4m

so the difference in distance is 295.3 but got it wrong!

what am i doing wrong..? and umm I'm not 100% sure why you have to add 24.92 twice... is it because it's symmtrical?