1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Distance covered on the moon

  1. Jan 22, 2014 #1
    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. jcsd
  3. Jan 22, 2014 #2
    Got it. I got the range formula wrong. the answer is c)
     
  4. Jan 22, 2014 #3

    BruceW

    User Avatar
    Homework Helper

    nice save :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Distance covered on the moon
  1. Moon exchange particle (Replies: 1)

Loading...