# Distance covered on the moon

1. Jan 22, 2014

### JonNash

1. The problem statement, all variables and given/known data

A stone is thrown with an initial velocity 5m/s such that it covers maximum possible horizontal distance RE on the surface of the Earth. The same stone is thrown with same initial velocity by a person standing on the moon, making an angle of 150 with the surface. given that acceleration due to gravity on earth's surface is g=9.8m/s2 and that on moon's surface is gm=1.6m/s2, the horizontal distnace it covers on the moon is
a) 0.326RE
b) 0.326/RE
c) 3.06RE
d) -0.629RE

2. Relevant equations

RE=u2sin2θ/g

3. The attempt at a solution

RE=u2/2ge
so u2=2geRE
Now, RM=u2sin215/gm
substitute u2 in RM
therefore RM=2geRE(0.067)/1.6
after solving I get RM=RE(0.82)
I know I'm missing the mark due to some stupid mistake somewhere, could you please point it out?

2. Jan 22, 2014

### JonNash

Got it. I got the range formula wrong. the answer is c)

3. Jan 22, 2014

nice save :)