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!

Velocity of a Projection In Projectile Motion

  1. Oct 24, 2012 #1
    1. The problem statement, all variables and given/known data

    For a given velocity of projection in a projectile motion, the maximum horizontal distance is possible only at ө = 45°. Substantiate your answer with mathematical support.


    2. Relevant equations
    My teacher gave us the information that u=10 m/sec, however I don't see how she arrived at that conclusion, also where did the angles of projection come from?



    3. The attempt at a solution
    if u = 10 m/sec , and angles of projection are 30o, 45o and 50o then,
    the horizontal distances travelled are
    R1 = 10^2 x sin 60 / 10 = 5 square root of 3 m
    R2= 10^2 x sin 90 / 10 = 10 m
    R3 = 10^2 x sin 100 / 10 = 10 x 0.9848 = 9.848 m
    thus, you see that the distance is maximum for 45o
     
  2. jcsd
  3. Oct 24, 2012 #2
    She probably wants you to derive an expression horizontal distance with an unknown θ, and then use your knowledge of trig functions to explain why the distance is maximized at θ=45°. I imagine that she gave you an initial velocity to work with simply so that you would have one less variable to be confused by, and obviously the θ you come up with will be independent of what the initial velocity is (so long as it's greater than 0).
     
  4. Oct 24, 2012 #3
    It looks like you have the right expression for the range of the projectile, R, R(theta, velocity) = you know what it is. Now take the derivative of R with respect to theta (velocity held fixed) and set that = to zero and solve for theta.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Velocity of a Projection In Projectile Motion
Loading...