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Projectile Motion Angle Problem

  1. Feb 1, 2013 #1
    1. The problem statement, all variables and given/known data

    A projectile is shot from the edge of a cliff (h=265m) above ground level with an initail speed of Vi= 115m/s at an angle of 37.0 degrees with the horizontal.

    a. Determine the time taken by the projectile to hit point P at ground level?
    b. Determine the Range of the projectile as meaured from the base of the cliff
    c. At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity.
    d. What is the magnitude of the velocity?
    e. What is the angle made by the velocity vector with the horizontal?
    f. Find the maximum height above the cliff top reached by the projectile.

    2. Relevant equations

    a. y=y 0 +v 0 t+(1/2)at 2
    b. R=(V o^2 Sin 2∅)/g
    c. ?
    d. ?
    e. ?
    f. ?


    3. The attempt at a solution

    Sorry if this seems to easy for you, but I'm 48 and my math skills aren't as sharp as they used to be. Im particualary stuck on trying to solve for the time. How do Solve for t? forgot the rules on how to factor out the t, so here is what I had:

    y=265m=(69.2087)(t)-(1/2)(9.8)(t)^2
     
  2. jcsd
  3. Feb 1, 2013 #2

    cepheid

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    Staff Emeritus
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    Gold Member

    Welcome to PF TRobison:

    When the projectile hits the ground, the vertical position is y = 0. So the equation becomes:

    0 = y0 + v0t - (1/2)gt2

    where y0 is the initial vertical position, which = h = 265 m

    v0 is the initial vertical speed, which you have to solve for using the total speed and launch angle, using trig (it looks like you already did this correctly, to get 69.2 m/s).

    This equation is a quadratic equation for t. It's not obviously factorable, so to solve it, you need to use the quadratic formula:

    http://en.wikipedia.org/wiki/Quadratic_equation#Quadratic_formula

    Any quadratic equation has two roots, which means that you will get two solutions using this formula: there will be TWO values of t for which y = 0, since the y vs. t curve is a parabola. However, only one of them will be physically meaningful. The other one will be negative, since if you extrapolate the parabolic trajectory to times before t = 0, the path will eventually intersect the ground again. Of course, this didn't actually happen, and times before t = 0 are not meaningful here.
     
  4. Feb 2, 2013 #3
    Thanks for putting me the right direction, I'm now calculating a total time of 17.05s my range seems too large, 1565.9m, but will continue with the rest of the problem to see how it comes out. Again thanks for your help.
     
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