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Speed at Maximum Height and Initial Velocity

  1. Oct 8, 2013 #1
    1. The problem statement, all variables and given/known data

    An ball is launched at a 45 degree angle with an initial speed v0. What is its speed at its max height?


    2. Relevant equations

    n/a

    3. The attempt at a solution

    This problem does not contain any calculations. I was wondering if somebody could explain how the speed at the max height is between v0 and v0/2.
     
  2. jcsd
  3. Oct 8, 2013 #2
    In projectile motion, horizontal velocity never changes. But vertical velocity keeps decreasing as the ball rises, and ultimately reaches zero at maximum height. So, essentially the velocity at max height is simply the horizontal velocity and you know that horizontal velocity never changes. I guess now you can figure it out.
     
  4. Oct 8, 2013 #3
    I think it would be easier for you to visualize if you were to draw a force diagram for the ball at the initial position, and at the maximum height position. When you draw these, it will be easy for you to see that after the initial launch, it's only gravity that acts upon the ball, (assuming no air friction). For this reason, it's very easy to split the velocity in terms of the x-direction and y-direction. The velocity in the x-direction remains constant, as there are no forces acting in that direction: while the velocity in the y-direction is being changed by the force of gravity. When the ball reaches the maximum height, the velocity in the y-direction is zero, and you're left with only the velocity in the x-direction. Use the angle you're given to solve for these two components.
     
  5. Oct 8, 2013 #4
  6. Oct 8, 2013 #5
    Wouldn't that mean that the speed at max height is v0*(cos45).
     
  7. Oct 8, 2013 #6
    Yes, that is correct. Since v0 is split by the angle of 45 degrees in its initial conditions, the x component, as you stated, is v0*(cos(45)), and the y component would be V0*(sin(45)). While the y velocity changes due to gravity, the x velocity remains the same.

    At max height, the y velocity is equal to 0, and the total velocity at that moment is just the velocity in x.

    With your answer, solving for cos(45), you can prove that your velocity will remain between v0 and v0/2 .
     
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