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Projectile Motion of a thrown stone

  1. Sep 24, 2007 #1
    1. The problem statement, all variables and given/known data
    #1
    A student stands at the edge of a cliff and throws a stone
    horizontally over the edge with a speed of 18 m/s. The cliff is 50 m above a flat horizontal beach. How long after being released does the stone strike the beach below the cliff ? With what speed and angle of impact does it land ?

    -----------------------------------------
    #2
    A science student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of 10 m/s. The student throws a ball along a path that she judges to make an initial angle of 60*, with the horizontal and to be in line with the track. The student's professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does the ball rise.


    2. The attempt at a solution
    #1
    Vx0 = 18 m/s
    d = rt --> x = 18t

    Dy = -50 m
    a = -9.8 m/s2
    -50 = (1/2)(-9.8)t2
    which t2 = negative, it is imposible



    #2
    Vx0 = 10 m/s

    Vx0 = 10 (cos 60*) = 5 m/s
    Vy0 = 10 (sin 60*) = 8.7 m/s

    Am I on the right track on both problem ? Can you guys help me ?
     
  2. jcsd
  3. Sep 24, 2007 #2

    learningphysics

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    t^2 isn't negative. What is t^2?

    There are two ways to do this... using energy (much easier)... or just using kinematics... have you covered energy yet in your class?
     
  4. Sep 24, 2007 #3
    #1
    Oh, It was a mistake !
    t = 3.2 s
    How can you find the speed and angle of impact does it land ?


    #2
    I think my teacher haven't covered energy yet !
     
  5. Sep 24, 2007 #4

    learningphysics

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    THe horizontal velocity remains 18m/s. What is the vertical velocity when impact happens?

    Ok... so suppose the ball is at the bottom of that 60 degree incline... what is the acceleration along the incline? First find the force along the incline, then you can get the acceleration along the incline...
     
  6. Sep 24, 2007 #5
    #1
    Vy = -160 m/s


    #2
    Is the picture below right ?
    10 m/s to the East, and then another line 30* N of E,
     
  7. Sep 24, 2007 #6

    learningphysics

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    Vy should just be 3.2*(-g) = 3.2(-9.8) right?

    Yeah... but I'm a little confused by how the question is asked... is the ball initially going at 10m/s up the 30 degree incline? They say it is in line with the track... does that mean that the ball's horizontal component is 10m/s... ie vcos(30) = 10m/s ?
     
  8. Sep 24, 2007 #7
    #1
    how to find the angle ?

    #2
    I am not sure about it too. They only said the train is running at constant 10 m/s, they didn't said about the velocity of the ball
     
    Last edited: Sep 24, 2007
  9. Sep 24, 2007 #8

    learningphysics

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    arctan(vy/vx)

    I'm guessing the ball starts at 10m/s... and it's going up an incline at 60degrees... ie E 60degrees N.

    So what is the net force acting on the ball along the plane.
     
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