Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Physics question - 2D movement please help!

  1. Mar 3, 2007 #1
    1. The problem statement, all variables and given/known data

    A rifle, in which the initial velocity of the bullet is 460m/s, is aimed at a small target 800 metres away at the same height as the rifle. At what angle above the ground must the rifle be pointed so that it hits the target?

    x = horizontal distance
    t = time
    Vx = horizontal velocity
    g = gravitational acceleration
    y = vertical distance
    Vyi = initial vertical velocity
    Vyf = final vertical velocity

    so far I know:
    x is 800m (given)
    g = -9.81m/s2, since the bullet is being pulled down by gravity
    y = 0, because the bullet does not travel any vertical distance because it lands at the same height that it is shot at.

    http://img144.imageshack.us/img144/1736/physicsproblem2bs1.jpg [Broken]

    2. Relevant equations

    y = Vyit + .5gt2
    y = Vyft - .5gt2
    g = (Vyf - Vyi)/t
    Vyf2 = Vyi2 + 2gy

    Vx = xt

    3. The attempt at a solution

    I'm almost clueless on how to approach this one. I can't even see how there is enough info given to be able to solve it. all I have so far is a diagram

    http://img464.imageshack.us/img464/4261/physicsproblemvf7.jpg [Broken]

    Thanks in advance!

    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 4, 2007 #2

    Very nicely laid out, a real treat to see a problem presented so legibly. Just for the sake of argument, suppose the bullet is fired horizontally, the time taken would be 800/460=1.74sec. The bullet would drop in that same time as you suggest .5*9.8*1.74^2=14.82m. In other words its a small angle indeed.
    Hint: know any approximations using small angles.

    If not, look at the first half of the flight to apex:
    we can solve for t in both the x any y axes. Knowing that Vy at apex = 0,
    then 460*sin (theta)=g*t Set that equal to the t for 400m along the x axis,

    should end up with an expression along lines of sin(theta)*cos(theta)=a
    where a is small. There is a trig identity that says:

    sin(2*theta)=2*sin(theta)*cos(theta). Again this is just one way to solve it, but there are others, esp if using small angle approxamitions like

  4. Mar 4, 2007 #3
    Thanks for the reply denverdoc. would you be able to clarify a few things?

    1.74s cannot be used for the time, since the bullet is being fired at an angle, right? Also I'm not familiar with small angle approximations, so maybe the first method is not the best way to go about.

    I understood "460*sin (theta)=g*t Set that equal to the t for 400m along the x axis", but you lost me on this part. Please explain how you got to sin(theta)*cos(theta)=a, and what did you mean by a?

  5. Mar 4, 2007 #4
    OK, two things happen simultaneously at apex. You hit the 1/2 way point on the x axis--this is 400m. You know the initial velocity and appreciate that not all of it is along the x axis. Also, You hit the condition of zero velocity along the Y. This leads to a separate eqn for t. Equate the two. a is just an arbitrary small number less than 1 which represents the constants all lumped together.
    Last edited: Mar 4, 2007
  6. Mar 4, 2007 #5
    You can find the solution with the rules that you already know. You just need to include a few more equations, more specifically the quadratic formula and the pythagorean theorem.
    Last edited: Mar 4, 2007
  7. Mar 5, 2007 #6
    Figured it out... I appreciate the help!


  8. Mar 5, 2007 #7
    glad to hear it. a good problem, well posted, and soluble thru a variety of methods.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook