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Homework Help: Projectile Motion in One Dimension

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data

    When a ball was thrown and caught at the same height from which it was thrown, it was measured to have travelled 12 meters in 1.3 seconds. What was the launch velocity?

    The answer on the key says 11.2 meters per second.

    2. Relevant equations

    Yfinal = Yinit + Vt + 1/2 a t^2

    3. The attempt at a solution

    If it goes 12 meters in 1.3 seconds, then it hit its peak in .65 seconds at 6 meters.

    So Yfinal = 6 and Yinit = 0 and t = .65 and a = -9.81

    Plugging that all in and solving for V I get 12.419 m/s. But my teacher on the key says the answer is 11.2 m/s. Where am I going wrong?
  2. jcsd
  3. Nov 8, 2012 #2
    vf2 = vi2 + 2ad

    looking at first half of the balls journey.

    vf = 0 m/s
    a = -9.8 m/s2 [up]
    d = 6 m

    Solve for vi

    vi = √(vf2 - 2ad)

    = √(-2ad)
    =10.84 m/s [up]

    getting a different answer :uhh:. I did not use time in my calculation though.
    Last edited: Nov 8, 2012
  4. Nov 8, 2012 #3
    I tried that method too, but that answer also wasn't an answer choice.
  5. Nov 9, 2012 #4


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    It's not as clear as it might be, but I suspect the question means that it travelled 12m horizontally.
  6. Nov 9, 2012 #5


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    Homework Helper

    The Title says this is Projectile Motion, so this projectile travels up for 0.65 sec, then comes down for 0.65. What speed does an object reach in 0.65 seconds [a little less than 6.5 m/s I suggest]
    Horizontally speaking - the projectile has travelled 12m in 1.3 seconds; that means a horizontal component of a little less than 10.

    Lets imagine those values were 6 and 10. Using Pythagoras gives v = √136 which is about 11.6 [note 136 = 62 + 102
    No doubt if you use the correct values you will get the 11.2 you are after.
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