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Projectile motion of a thrown rock

  1. Feb 6, 2012 #1
    1. The problem statement, all variables and given/known data

    A hiker throws a stone from the upper edge of a vertical cliff. The stone ' s initial velocity is 25.0 m/s directed at 40.0 degrees with the face of the cliff, as shown in Fig. 3.1. The stone hits the ground 3.75 s after being thrown and feels no appreciable air resistance as it falls. the height of the cliff is closest to

    Here is the link to the picture: http://img10.imageshack.us/img10/8058/58525486.jpg [Broken]

    2. Relevant equations



    3. The attempt at a solution

    So what I first did was break the velocity into its vertical and horizontal components,

    25.0sin40=16.06m/s(y0)

    25.0cos40=19.15m/s(x0)

    We know the time it takes to hit the gound is 3.75 seconds

    I got its Vy( final velocity) by using the following equation vy=vosina-gt

    vy=16.0sin40-9.8(3.75s)
    =-26.46m/s

    Here is where I start getting lost, I wanted to solve for deltay using the following formula;

    vf^2=vi^2+2aDeltay

    (-24.46)^2-(16.06)^2/2(-9.8).. and I got some funky answer, the answer to the problem is 141.

    Can someone walk me through how to get that answer, thank you!
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 6, 2012 #2
    Well to start off 25sin40 is the x component of the velocity, not the y, assuming theta is the angle between the vertical cliff face and the direction the stone was thrown, although I could just be reading it wrong, the picture is open to interpretation...

    Your kinematics equations are fine, it looks like a simple error in your algebra as long as the above error is irrelevant.

    Additionally, what are you solving for?
     
  4. Feb 6, 2012 #3
    Sorry I edited it, we are solving for the height of the cliff.
     
  5. Feb 6, 2012 #4
    Check your algebra, and trig. Again, I don't think your angle is in the right spot. I solved it and got the correct answer.

    Your kinematics equations themselves are fine.
     
  6. Feb 6, 2012 #5
    I tried 25cos40=v0y=19.15

    y=19.15(3.75)-1/2(-9.8)(3.75)^2=141!!!!

    Thank you!!

    Can You walk me through your thought process when you were solving this problem?
     
  7. Feb 6, 2012 #6
    Sure :)

    Ok so when I see a kinematics equation the first thing I do is resolve it into components, just as you did. I find that

    v0x = 25sin40
    v0y = 25cos40

    Since we're concerned only with the height of the cliff, and we know the time it takes for it to fall, it becomes quite simple. You can solve one of two ways:

    y = y0 + v0yt + .5at2

    or solve for the final velocity in the y direction, and then use the time independent kinematics equation (thats the way we both did it, it just occurred to me that the way above works as well, and in one less step!)

    Then just solve for x (making sure that your x0 correctly represents the initial position.

    Hope this helps! :)
     
  8. Feb 6, 2012 #7
    Clear explanation, thank you!
     
  9. Feb 6, 2012 #8
    No problem.
     
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