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Simple Laws of motion question

  1. Oct 20, 2006 #1
    A friend and I were conducting a science experiment, where an object weighing roughly 550 grams was launched vertically, and after examining the video footage, was airborne for aproximatly 4.5 secconds. Having never studied physics before, I was curious if you could calculate the launch speed and or peak height from this information.

    Any help would be highly appreciated.
     
    Last edited: Oct 20, 2006
  2. jcsd
  3. Oct 20, 2006 #2

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    Assuming negligible air resistence, wich is approximatelytrue if the object is a ball made of a dense material, the velocity of a body moving against gravity is [tex]v = v_0 - gt[/tex]
    At the top of the trajectory the velocity is 0 and the time is half the total time.
     
  4. Oct 20, 2006 #3
    I don't really understand that sorry, but according to Wikipedia, a stationary object's speed increases by 9.81 metres per second as it falls. and since my object was falling for 4.5/2, 2.25 secconds, I figured the peak hieght would be 9.81 + (9.81x2) + (9.81x3)/4 which equals 36.7875 metres, which, having seen the object, sounds about right. And 9.81 being approximatly 22 miles per hour, I guessed the speed it was traveling when it hit the ground and indeed the speed it was launched at would be 22 x 2.25 which is 49.5 miles per hour.
    However, having no expirience with physics, I am hesitant to trust these results. Could anyone confirm or correct this?
    Thank you for your input.
     
  5. Oct 20, 2006 #4

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    I don't understand the calculations you made. The initial and final velocities are equal in absolute values.
    [tex]v_0 = 9.81*2.25 = 22m/s = 11.4 mph[/tex]
    The maximum heogth will be
    [tex]h = v_0*t - \frac{1}{2]g*t^2 = 24.8m[/tex]
     
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