1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Projectile motion particle question

  1. Apr 7, 2005 #1
    A particle is projected from the origin 0 with initial velocity v at an angle of elevation [tex] \theta [/tex] and it moves freely under gravity. Find [tex] tan \theta [/tex] if the greatest height reached by the particle is equals to its range on the horizontal plane.

    Using the equations of motion, i have found

    [tex] \frac {v^2_y}{ 2g} = v_x t =v_y - 1/2 gt^2 [/tex]

    but i am unsure of how to manipulate it from here to find [tex] v_y / v_x [/tex].
  2. jcsd
  3. Apr 7, 2005 #2
    Use these

    [tex]v_x = vcos( \theta)[/tex]
    [tex]v_y = vsin( \theta) -gt[/tex]

    [tex]x = vcos (\theta)t[/tex]
    [tex]y = vsin( \theta)t - g \frac{t^2}{2} [/tex]

    Then, the maximal heigth can be determined by realizing that v_y must be zero there. Calculate the time at which this occurs and put it into the formula for y. same goes for x (set y = 0 and calculate the time and put it into x) and then set these two equal to each other

    Last edited: Apr 7, 2005
  4. Apr 9, 2005 #3
    hey thanks for the help marlon, I've gotten it....
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook