Distance covered on the moon

[JonNash]

Homework Statement

A stone is thrown with an initial velocity 5m/s such that it covers maximum possible horizontal distance R_E 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 g_m=1.6m/s², the horizontal distnace it covers on the moon is
a) 0.326R_E
b) 0.326/R_E
c) 3.06R_E
d) -0.629R_E

Homework Equations

R_E=u²sin²θ/g

The Attempt at a Solution

R_E=u²/2g_e
so u²=2g_eR_E
Now, R_M=u²sin²15/g_m
substitute u² in R_M
therefore R_M=2g_eR_E(0.067)/1.6
after solving I get R_M=R_E(0.82)
I know I'm missing the mark due to some stupid mistake somewhere, could you please point it out?

 

[JonNash]

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

 

[BruceW]

nice save :)