Notices
Results 1 to 10 of 10

Thread: Straight Line Kinematics

  1. #1 Straight Line Kinematics 
    Forum Freshman
    Join Date
    Sep 2006
    Posts
    32
    A ball is thrown vertically upwards. It rises, falls back down, strikes the floor and bounces back up hwere it is caught at the same position of that it was originally released. Assume that no energy is lost when the ball hits the floor.

    If anyone can help the aforementioned is to be graphed, but by no means am I asking you to do it for me. The problem is that the graph has to have a scale. Now, I figure a velocity time graph would start above the time axis, to come in contact with the time axis, to go below the time axis where it would increase in velocity due to acceleration and longer distance, to return back to the time axis (pulsating after hiting the ground), and then would end above the time axis. I am also given the initial speed (10m/s), where it is released (2m above the ground), and the acceleration is assumed to be 10m/s^2 [down]. If anyone can point in the right direction in order to make a scale for this graph...it would be greatly appreciated as I have entered this course with no prerequistes and am having great difficulty making sense of things. thanks.


    Reply With Quote  
     

  2.  
     

  3. #2  
    Forum Freshman FancyShoes's Avatar
    Join Date
    Mar 2007
    Posts
    34
    Well I do not know if this will help you out that much but on the y axis would be height and the x axis would be time. So you would plot down the highest point the mid point and the lowest point. I am guessing that it would look like a sine curve since you are throwing the ball up from the mid point and it is increasing then it is falling then going down through the mid point to the lowest point then bouncing back up to the mid point. I AM NOT 100% SURE. So wait for other comments as well because I might not be right :/.


    whats the interval of 1/mower? lnmower... heh
    Reply With Quote  
     

  4. #3  
    Forum Freshman
    Join Date
    Mar 2007
    Location
    Canada
    Posts
    34
    If I'm correct, the velocity should be in the Y direction and time in the X direction. Since we're dealing with the vertical direction only, as the ball accelerates it increases in velocity, when it hits the ground it's at it's peak velocity.. when it bounces back up it has a negative acceleration (because it's going against gravity) and thus, by the time it comes to it's original height, the final velocity should be zero.

    I'm assuming you're asked to ignore air resistance and elasticity of the bounce/collision.
    Reply With Quote  
     

  5. #4  
    Forum Freshman FancyShoes's Avatar
    Join Date
    Mar 2007
    Posts
    34
    would it not have a negative velocity going down? time up=time down right? but time down is negative?
    whats the interval of 1/mower? lnmower... heh
    Reply With Quote  
     

  6. #5  
    Forum Freshman
    Join Date
    Mar 2007
    Location
    Canada
    Posts
    34
    When the velocity is negative it's simply stating what direction it's moving in. If you consider the postive Y to be "up" then as the ball falls, the velocity will be negative.
    Reply With Quote  
     

  7. #6 Re: Straight Line Kinematics 
    Forum Sophomore basim's Avatar
    Join Date
    Mar 2007
    Location
    maldives
    Posts
    138
    Quote Originally Posted by ender7x77
    A ball is thrown vertically upwards. It rises, falls back down, strikes the floor and bounces back up hwere it is caught at the same position of that it was originally released. Assume that no energy is lost when the ball hits the floor.

    If anyone can help the aforementioned is to be graphed, but by no means am I asking you to do it for me. The problem is that the graph has to have a scale. Now, I figure a velocity time graph would start above the time axis, to come in contact with the time axis, to go below the time axis where it would increase in velocity due to acceleration and longer distance, to return back to the time axis (pulsating after hiting the ground), and then would end above the time axis. I am also given the initial speed (10m/s), where it is released (2m above the ground), and the acceleration is assumed to be 10m/s^2 [down]. If anyone can point in the right direction in order to make a scale for this graph...it would be greatly appreciated as I have entered this course with no prerequistes and am having great difficulty making sense of things. thanks.
    can you exactly mension what is your problem? i think your way of questioning is comfusing :wink:
    Reply With Quote  
     

  8. #7  
    Forum Freshman wonkothesane's Avatar
    Join Date
    Feb 2007
    Location
    Brisbane, Australia
    Posts
    47
    Well I do not know if this will help you out that much but on the y axis would be height and the x axis would be time. So you would plot down the highest point the mid point and the lowest point. I am guessing that it would look like a sine curve since you are throwing the ball up from the mid point and it is increasing then it is falling then going down through the mid point to the lowest point then bouncing back up to the mid point. I AM NOT 100% SURE. So wait for other comments as well because I might not be right :/.
    Not really a sine curve. If the acceleration due to gravity is constant in magnitude (only changes direction at ground and highest point) then the derivative gives you the velocity (ie v = da/dt = at) so you would get a triangular wave for velocity. Furthermore, position is the derivative of velocity so height = dv/dt = d(at)/dt = (1/2)*at^2. This is a parobola so cos the ball is bouncing with no energy loss you would get a periodic parobala (whatever that is called).

    Note that the above is basically one solution to Newton's F = ma, though I didn't use any initial conditions.
    Fry me a kipper skipper, I'll be back for breakfast!
    Reply With Quote  
     

  9. #8  
    Forum Senior anand_kapadia's Avatar
    Join Date
    May 2006
    Location
    India
    Posts
    300
    Quote Originally Posted by ender7x77
    A ball is thrown vertically upwards. It rises, falls back down, strikes the floor and bounces back up hwere it is caught at the same position of that it was originally released. Assume that no energy is lost when the ball hits the floor.
    I didn't understand what you said. No energy lost when it hits the floor. It is the inertia of motion that causes it to bounce.
    Correct me if I am wrong.
    Reply With Quote  
     

  10. #9  
    Forum Sophomore basim's Avatar
    Join Date
    Mar 2007
    Location
    maldives
    Posts
    138
    it is the kinetic energy of the ball which cause it to bounce.
    when it hits the ground the ball bounces with the same velocity, if the collision is perfectly elastic.
    Reply With Quote  
     

  11. #10  
    Forum Senior anand_kapadia's Avatar
    Join Date
    May 2006
    Location
    India
    Posts
    300
    Ohh right.
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •