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Help, Gravity on objects thrown up

  1. Mar 13, 2009 #1
    In the 7th grade we don't learn much if any physics, so in the past few years I've been learning quantum mechanics and relativity (without the mathematics). due to the fact that there is a lack of mathematics being thought at this age i'm really not into classical physics. but i do have a question if i threw a 1kg ball in the air at a constant velocity of 10 meters a second (neglecting the acceleration in the first place) how will i know when the object will stop going up and start heading down (done at the surface of the earth so i know it'll fall at a rate of 9.8 m/s2.) and no air resistance. Help!
     
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  3. Mar 13, 2009 #2

    cristo

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    What do you mean you're "really not into classical physics"? Whilst it's admirable that you've been reading about quantum mechanics and relativity, you really won't be able to fully understand the subjects until you have a (mathematical) grasp of classical mechanics. Thus, I would urge you to try and learn this now!

    Since you seem genuine, though, and not like this is a homework question, I'll help you out. Perhaps you have heard of the kinematic equations (or whatever name they are commonly presented under in your education system). They are listed here: http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/1DKin/U1L6a.html [Broken] These equations enable us to consider uniform acceleration in systems without air resistance.

    The third equation looks like this [itex]v_f=v_i+at[/itex], where v_f and v_i are the final and initial velocity, respectively, a is acceleration of the ball and t is time. When you throw a ball up in the air, it gets decelerated by the gravitational field of the earth. At the point where the ball stops going up and starts going down, the ball will have zero velocity; i.e. as the ball travels upward, its velocity starts positive, decreases to zero, and then 'decreases' so it is negative.. the ball is falling downwards.

    Thus, taking the positive direction as up, your equation will read [itex]0=10-9.8*t[/itex]. That is, the ball will reach its maximum height at around 1.02 seconds after release.
     
    Last edited by a moderator: May 4, 2017
  4. Mar 13, 2009 #3
    You can use
    [tex]x(t)=x_0 + v_0t+\frac{1}{2}at^2[/tex]
    and conservation of energy
    [tex]\frac{1}{2}mv^2 = mgh[/tex]
    You could also extremize the first equation, but this way gets you the answer, too. We can solve for h, so that we know how high the ball gets (relative to the point where we let the ball go).
    [tex]h = \frac{v_0^2}{2g}[/tex].
    Since we are doing things relative to our hand, for some time T the first equation will be
    [tex]\frac{v_0^2}{2g} = v_0T - \frac{1}{2}gT^2[/tex]
    Solve for T.
     
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