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Projectile Motion Problem

  1. Feb 5, 2016 #1
    1. The problem statement, all variables and given/known data
    I saw this in a video about solving projectile motion problems and I was confused, I think I may have seen the same thing in my textbook:


    The question says: A soccer player on a level playing field kicks a soccer ball with a velocity of 9.4 m/s at an angle of 40 degrees above the horizontal. Determine the soccer ball's maximum height.

    vi=6.042m/s, a=-9.8m/s, vf=o m/s

    2. Relevant equations

    vf^2=vi^2+2a(displacement)
    3. The attempt at a solutIon
    I saw this in a video and I wasn't trying to solve it myself. But, I was confused because the initial velocity used was calculated previously as the initial velocity for the y component of the motion since this is projectile motion and it has 2 components. The final velocity is the velocity at the maximum height- om/s. But, the thing that confused me was that acceleration was stated to be -9.8m/s^2. If this question is taking the initial velocity from when the ball was thrown and the final velocity at the point at which the ball is at its maximum height, is it not then looking at the ball's motion from when it is thrown up until it reaches it's maximum height only- it's looking at the ball's motion as it is moving upwards. So, why is the acceleration given as if the ball were in free fall because that is the acceleration used in problems where an object is in free fall. Sorry this is so long, I just wanted to make sure that what I was saying made sense. Could someone please explain this to me, I'm really confused.
     
  2. jcsd
  3. Feb 5, 2016 #2
    You're neglecting air resistance and what not in such projectile motion problems. So the only force acting on the soccer ball is the force of gravity. It is the only source of acceleration. So it is taken as -9.8m/s^2. The - sign is because they took the upward direction to be positive. You could take velocity in the negative direction to be positive and acceleration would be 9.8m/s^2 but your end answer would be the same.
     
  4. Feb 5, 2016 #3
    How is the ball acceleratng upwards at 9.8m/s^2 when accelleration is equal to vf-vi, which would be -6.042m/s^2? And for this problem, air resistance is not taken into account because this is introductory physics.
     
  5. Feb 5, 2016 #4

    cnh1995

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    Are you sure this is acceleration?
     
  6. Feb 5, 2016 #5
    Sorry Sheldon, acceleration is equal to change in velocity over time, it should be -6.042/time, and the time for the object to complete its entire motion is 1.2 seconds- I think you may be able to divide that by 2 to get the time for it to move from its starting position to its maximum height, but I'm not sure because objects do accelerate as they fall so it's motion from the maximum height tp the ground may take a shorter amount of time than its motion from the ground to the maximum height. In which case, I think it would be wrong to divide the time by 2 and say that it takes an equal amount of time for the ball to move from the ground to the max height as it takes for the ball to move from the max height to the ground.
     
  7. Feb 5, 2016 #6

    SteamKing

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    Who said the ball was accelerating upward? An upward velocity is taken to be positive, and acceleration due to gravity is taken to be -9.8 m/s2, which means that while the initial velocity of the kicked ball is upward, that velocity is decreasing while it is aloft. Eventually, the velocity of the ball upward reduces to zero, after which, the ball falls back to the ground.

    You should be careful with units here. Adding to subtracting like units results in like units.
     
  8. Feb 5, 2016 #7
    I'm still confused- I thought an acceleration of -9.8m/s^2 was limited to objects in free fall
     
  9. Feb 5, 2016 #8

    cnh1995

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    When a body is moving with some velocity v and a force is applied on it, component of velocity along the line of action of force changes. This means, along the direction of force, body has acceleration. Here, gravity is in acting in y direction, hence, y component of velocity is affected. Hence, g=9.8m/s2 is applicable here for the y component of the velocity.
     
  10. Feb 5, 2016 #9
    But, it's not falling down- it's moving upwards. I don't understand.
     
  11. Feb 5, 2016 #10
    "Sheldon" is right mate. What you have taken as vf is not actually vf at all. It is only the vertical component of vf. Look at it this way: Is there any force acting in the horizontal direction to bring about a change in the horizontal component of the velocity of the ball? If there isn't why should that horizontal component change?
     
  12. Feb 5, 2016 #11

    cnh1995

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    Why does a ball thrown vertically upwards stop at some point? What affects its upward journey?
     
  13. Feb 5, 2016 #12
    A downward acceleration does not necessarily mean that the object itself must move downwards. Look at the vertical component of the velocity of the ball. It is decreasing, so there must be a force acting in the opposite direction to the direction of the vertical component. The ball moves upwards with decreasing velocity so there must be a force and hence an acceleration acting in the opposite direction.
    Also the ball does eventually start falling down. After it reaches its maximum, it
     
  14. Feb 5, 2016 #13
    But we're only looking at this in terms of the y component because we want to find its max height, which has to do with the y component, not the x component
     
  15. Feb 5, 2016 #14

    ehild

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    The ball performs a 2-dimensional motion, it moves both vertically and horizontally. If you set up a coordinate system with horizontal x axis pointing to the right and vertical y axis pointing upward, then both components of the initial velocity are positive. The acceleration points vertically downward during the whole motion, so ay=-9.81 m/s2 and ax=0.
     
  16. Feb 5, 2016 #15
    Gravity- so even as in object is travelling upwards, it's acceleration is still -9.8m/s^2?
     
  17. Feb 5, 2016 #16
    So even as an object is moving upwards freely, its acceleration is still -9.8m/s^2?
     
  18. Feb 5, 2016 #17

    cnh1995

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    Right. The ball will be decelerating, hence the minus sign.
     
  19. Feb 5, 2016 #18
    Yes. The reason I mentioned the horizontal component was because I was not entirely sure you understood that the acceleration in this case will be independent of the horizontal components of velocity. Also change in velocity is not acceleration which was the point "Sheldon" was trying to make.
     
  20. Feb 5, 2016 #19
    Okay, makes sense. Thank you.
     
  21. Feb 5, 2016 #20
    Incidentally watch one of Walter Lewin's lectures on youtube. There is one on projectile motion. It will really clear things up for you.
     
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