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Homework Help: Gravity, velocity time and distance problem

  1. Feb 14, 2012 #1
    1. The problem statement, all variables and given/known data
    A rock from high of ##h=180m## goes down and the gravity is ##g=9.81\frac{m}{s^2}##.
    Find for how much time did the rock fall down and the velocity of the rock.


    2. Relevant equations
    ##v^2=2gh##
    ##v=\sqrt{2gh}##
    ##t=\frac{v}{g}##


    3. The attempt at a solution
    First of all we will find the velocity because, to find time we need the velocity so:
    ##v=\sqrt{2gh}=\sqrt{2χ9.81\frac{m}{s^2}χ180m}=

    \sqrt{3531.6}=59.43 \frac{m}{s}##
    ##t=\frac{v}{g}=\frac{59.43\frac{m}{s}}{9.81\frac{m}{s^2}}=6.1s##
    is that correct way to do it?
    BTW. how to make the square root on formulas?
     
    Last edited: Feb 14, 2012
  2. jcsd
  3. Feb 14, 2012 #2
    What you have done is correct. Be careful with your significant figures. You provide 4 significant figures for the velocity yet only use 3 for g.

    I don't use formulas so I cannot help you there.
     
  4. Feb 14, 2012 #3
    Thanks, but i have two more questions:
    Where do we know that ##t=\frac{v}{g}##, if in formula of velocity ##v=√2gh## we don't have time ##t##?
    and would gravity formula be ##g=\frac{v^2}{h}## or ##g=\frac{v}{t}## and ##v=gt##?
     
  5. Feb 14, 2012 #4
    The parameter g is the acceleration of gravity in a free fall. In other words if you drop a ball, it accelerates toward the ground at a rate of 9.81 m/sec^2. The velocity attained during a free fall is (with no air drag)

    v = gt

    Rearranging,

    t=v/g


    As for v=sqrt(2gh), that formula can get derived a least a couple of ways.

    For an object dropped with no initial velocity, the distance it falls is

    h = 0.5 g t^2

    But we know that

    v = g t

    so t = v/g

    Substitute for t in the displacement equation

    h = 0.5 g (v/g)^2 = 0.5 v^2/g

    Therefore

    v = sqrt(2gh)

    The same formula can be derived from conservation of energy considerations.
     
  6. Feb 14, 2012 #5
    got it now thanks!
     
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