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Projectile Motion Question (physics 101)

  1. Sep 11, 2008 #1
    1. The problem statement, all variables and given/known data

    A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 65.0 m/s at an angle of 37 degrrees with the horizontal. Find the horizontal and vertical components of its velocity at the instant just before the projectile hits point P (the ground).

    2. Relevant equations

    Vxo = Vo cos 37
    Vyo = Vo sin 37
    Vy = Vyo - gt ...????

    3. The attempt at a solution

    The answer is given in the back of the book. Plugging in the initial velocity in the first equation gives the correct answer (which I expected), but the answer I acheive for Vyo should be negative and I haven't yet acheived that. I tried to plug the answer I got for Vyo (the positive one) into the equation Vy = Vyo - gt, but still did not get the right answer. I'm truly grasping at straws here.
  2. jcsd
  3. Sep 11, 2008 #2
    You can use Vy = Vyo - gt to find the vertical component, but must first find the time at which the projectile reaches P. To do this, you use the equation for the y-coordinate as a function of time.
  4. Sep 11, 2008 #3


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    Hi cheese89! :smile:

    I don't understand how you got any result for Vy from Vy = Vyo - gt … :confused:

    you don't know what t is.

    Hint: you haven't used y = 125 yet, have you? :wink:
  5. Sep 11, 2008 #4

    Its an oral question dear. Just think will the boy acclerate in x direction. If it does, vertical component will change otherwise not. For the y direction, whatever you throw come down with same speed.
  6. Sep 11, 2008 #5
    For this problem you should check your constant acceleration formulas. The Acceleration in this case is a constant -9.8m/s^2(gravity). And there is no acceleration in the x direction. The velocity would be v=Vo+at. where v= final velocity, Vo= initial velocity, a=acceleration, and t= time. Look at the information given to you in the problem and consult some of the other formulas. You have a displacement of 125m, a Vo, the acceleration, the angle. all you need is the time it took to get to the ground and you will have your answer =)
  7. Sep 11, 2008 #6


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    Hi ritwik06! :smile:

    That only applies when we throw from the ground … this is from a 125 m cliff. :wink:
  8. Sep 11, 2008 #7
    you have to find time first using the formula Y = vt + 1/2 g t^2
  9. Sep 12, 2008 #8
    I know. I thought cheese was working for that interval as well.
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