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One-Dimensional Motion

  1. Jun 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A projectile is fired horizontally from a height of 20 meters above the ground, with an initial velocity of
    7.0 m/s. How far does the projectile travel horizontally before it reaches the ground?

    2. Relevant equations
    X=VoT+1/2AT^2

    3. The attempt at a solution
    I solved for time, then got my answer by using d=rt.
    My answers: time=2.8 seconds and distance traveled=19.6m

    Their answer is 14m. Can someone explain where I went wrong?
     
  2. jcsd
  3. Jun 19, 2015 #2

    Lok

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    Could you give the complete solution or solving pathway chosen?
     
  4. Jun 19, 2015 #3
    They solved for time first, as I did, using the equation for position. X=VoT+1/2AT^2.
    Their equation is: -20=0t+1/2(-9.8)t^2. When they solved for t they got t=2 seconds.

    This is where I am confused because the equation states initial velocity is 7.0m/s, yet they use 0. That's where I need help.
     
  5. Jun 19, 2015 #4

    Lok

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    Well in which direction are you solving for time? And how would that translate in your equation?
     
  6. Jun 19, 2015 #5
    We are using the positive x-axis for the direction of time. They got +2 seconds, I'm getting +2.8 seconds. Does that clear it up? I'm kind of confused so I apologize if I'm not making sense.
     
  7. Jun 19, 2015 #6
    Image dropping another projectile from rest at the same time, they would both strike the ground a the same time.
    Ignore the horizontal component of motion for now ( a constant 7 m/s)
    Regardless of the horizontal velocity, the time taken to fall the 20 m to ground is the same.
    The projectile accelerates (vertically) from 0 m/s
    g is +ve as it acts in the direction of motion, so
    u = 0 m/s
    g = 9.8 (m/s)/s
    s = 20 m
    Use one of newtons rules to calculate the time t to fall the 20 m from rest
     
  8. Jun 19, 2015 #7

    Lok

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    Supposing the X axis is horizontal, your gravity is not present in the X axis, only the initial velocity, so solving for time there would give no ideea of when the projectile hits the ground. But it would in the Y, you have gravity, and the initial velocity in a vertical Y direction would be?
     
  9. Jun 19, 2015 #8
    This is what I'm not understanding. If the initial velocity is actually 0 m/s, then why would the equation state it is 7.0 m/s.
     
  10. Jun 19, 2015 #9
    Respond tommorow
     
  11. Jun 19, 2015 #10

    Lok

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    The point is your initial velocity is in the horizontal direction and gravity is vertical. Your equation applies only in a one dimensional setup so on a line, if you have conditions pointing in another direction than your equations dimension would permit you have to do a transformation with angles and sin and cos (usually at this point in your physics study). In your problem you have basically 2 dimensions. X horizontal and Y vertical. The equation in Y has only gravity and a 0 initial velocity, and in the X, zero gravity and your initial velocity of 7m/s. It's the same equation used in two different directions there the velocity and force vectors point at an angle of 90 deg.
     
  12. Jun 19, 2015 #11
    I understand what you are saying. I've only studied one-dimensional motion and (believe it or not) I haven't learned about vectors or two-dimensional motion at all. Would this problem then require more than just knowledge of one dimensional motion?
     
  13. Jun 19, 2015 #12

    Lok

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    It would be close to the need of such notions. And if not studied do ask your teacher about these and your problem.

    But just to clarify the problem somehow visually (with the coolest most powerfull GPU of imagination)
    When viewed from the far left (or right). The projectile will look like standing still at start and accelerating towards the ground with g until it hits it.
    When viewed from the far top, your projectile will be looking like it is moving at a constant speed from start, until for some reason it comes to a complete stop.
    The two motions are completely independent until the ground is hit, just as they should appear in the math.
     
  14. Jun 19, 2015 #13
    Hard to believe I would get confused with something so simple. You are awesome, and I appreciate you helping me understand!
     
  15. Jun 20, 2015 #14
    Splitting the motion into its horizontal (constant) and vertical (obeys newton's rules) components is fundamental to solving these types of problems, glad you got it.
     
  16. Jun 20, 2015 #15
    Can confirm that the solutions ( with g @ 9.8 (m/s)/s ) :
    t = 2.0203 s
    horizontal distance = 14.142 m
     
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