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!

Deriving Equations

  1. Sep 12, 2007 #1
    How do you derive the equation for the launch angle using the range and peak equations? This isn't a homework problem but something i want to know for my own general knowledge for solving physics problems. Is there a site that has physics equations solved for different variables? Hyperphysics is a great site but it doesn't work through the deriving and solving of the equations.

    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Sep 12, 2007 #2


    User Avatar

    Given its height, you can find the upward component of its initial velocity, and you can also find the time it is in flight, since you know the constant acceleration g. Given its range, and having found the time interval for which the projectile was in flight, you can find the horizontal component of its velocity. Having found those two componemts, the tangent of the angle is the ratio of the vertical component by the horizontal component of the velocity.
  4. Sep 12, 2007 #3
    how so if you have the y = yo + voyt - 1/2gt^2?, you'll have two unknown voy and t
  5. Sep 12, 2007 #4
    voy is your initial velocity. The problem MUST have some indication of velocity on the y axis otherwise, it would be unsolvable.
  6. Sep 12, 2007 #5
    A problem that I'm working on has a range of 25m and a height or peak of 4.90m. It says find the initial velocity,the angle at which the projectile is fired, and the time its in the air. I looked on hyperphysics and it said if range and peak are given an equation can be derived to solve for the angle.
  7. Sep 12, 2007 #6
    Ooo OK.

    We know the total travelled distance is 25m

    And the max height is 4.9m

    The equation for the velocities are

    [tex]v_x = v_0cos(\theta)[/tex]
    [tex]v_y = v_0sin(\theta)-gt[/tex]

    And for distances are

    [tex]s_x = v_0cos(\theta)t[/tex]
    [tex]s_y = v_0sin(\theta)t-\frac{1}{2}gt^2[/tex]

    First of all, plug 25m into the [tex]s_x[/tex] equation, solve for t to find the expression for
    the total time of travel.

    Then, plug in t into [tex]s_y[/tex] you still have [tex]v_0[/tex], [tex]theta[/tex] left as unknowns. Apparently, the y distance has to equal to 0 at this particular time you have found. Solve for [tex]\theta[/tex]. Now you have the derived expression for [tex]\theta[/tex].

    To find the actual values, you need the last piece of information, where the max height is 4.9m. It takes half of the time of total distance travel for the projectile to reach this max height. So plug in 4.9m as [tex]s_y[/tex], plug in t/2 into the expression, then it's just simultaneous equation - 2 equations and 2 unknowns
    Last edited: Sep 12, 2007
  8. Sep 12, 2007 #7


    User Avatar
    Homework Helper

    l46kok's explanation is excellent. Only thing I'd recommend is not plugging in actual numbers until the very end... meaning get the formula for arbitrary range R, and height H...

    Then once you have the equations for velocity and angle... then plug in the actual numbers...
  9. Sep 12, 2007 #8
    thanks for the explanation
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Deriving Equations
  1. Deriving an equation (Replies: 1)

  2. Deriving an equation? (Replies: 3)

  3. Derive the equation (Replies: 13)

  4. Derive an equation (Replies: 3)