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Projectile Motion's relationship with Kinetic Energy and Potential Energy

  1. True

    4 vote(s)
  2. False

    0 vote(s)
  1. Oct 7, 2011 #1
    Imagine a ramp setup on top of a tall table. The height Δy is measured. To find the initial velocity at the instant the ball leaves the ramp, I set up the kinetic energy and potential energy equal to each other to find the initial velocity of the x component.

    PE = KE
    m*g*(hr) = 0.5*m*v^2

    where hr is the height of the ramp and v is initial velocity (x-component)

    solving for vx (x-component velocity), I got:
    vx = √(2*g*hr)

    To get the time for the object's time in flight:
    y'-y= vy + 0.5gt^2
    Δy= vy + 0.5gt^2, where Δy is the height from the ground to the ramp.
    since θ= 0° I found t to be:
    t = √{ (2*∆y)/g }

    Now my question is how do I find the range of this object?
    I started out with Δx = vx*t ; where vx is the initial x-component velocity.... is that even right?
    I'm hesitant to use it because written as Δx/t, it looks like an average velocity equation.
    Furthermore, in Wikipedia I saw the equation

    d= {v*cos([itex]\Theta[/itex])}/g * [v*sin([itex]\Theta[/itex]) + sqrt(v*sin([itex]\Theta[/itex])^2+2g*y)]

    http://en.wikipedia.org/wiki/Range_of_a_projectile" [Broken]
    under uneven ground

    but the problem is I don't have final velocity................ or can I calculate the final velocity with the givens........ if so how??

    Could anyone please nudge me in the right direction to find Δx?

    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 7, 2011 #2
    I think I solved it........... duh.

    in the distance formula
    d = v(initial)*t + 1/2*a*t^2.

    v(initial) and time is already attained and the the a acceleration is -9.81m/s^2

    Correct me if I'm wrong.
    Last edited: Oct 7, 2011
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