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FreeFall motion Assumptions in proving

  1. Oct 10, 2012 #1
    1. The problem statement, all variables and given/known data
    a stone is thrown vertically upwards with a speed of u metres per second. a second stone is thrown vertically uppwards from the same point with the same initial speed but T seconds later than the first. prove that they collide at a distance (4u^2-g^2T^2)/(8g) metres above the point of projection.


    2. Relevant equations
    s=ut+.5(g)(t)^2


    3. The attempt at a solution
    now i should resolve vertically so i would get s= uT + .5(g)T^2 for the second and s= u(X-T)+ .5(g)(X-T)^2 for the first but when i attempt to get the answer i fail miserably or everything just cancels out. can someone help me with this?
    1. The problem statement, all variables and given/known data
    initial velocity is u and that time after second stone is thrown is T
     
  2. jcsd
  3. Oct 10, 2012 #2
    Can you show us what you have done so far? And which way is your coordinate axis pointing?
     
  4. Oct 10, 2012 #3
    well i got u(X-T)+ .5(g)(X-T)^2= uT + .5(g)T^2 thus it is equal to uX-2uT+.5gx^2-.5gXT thats all i could get
     
  5. Oct 10, 2012 #4
    what do you mean with coordinate axis?
     
  6. Oct 10, 2012 #5
    I mean, in which direction does the y axis point? Does it point upward? I.e. does higher positions correspond to higher values of y? Because, in that case the formula the gravitational acceleration is opposite to the y axis, and the acceleration [itex]a=-g[/itex]. You would then have
    [itex]y_1 = ut - \frac{gt^2}{2}[/itex]
    for the first stone, where [itex]t[/itex] is the time, and we let [itex]y=0[/itex] at the starting point of the throw. The other stone satisfies
    [itex]y_2 = u (t-T) - \frac{g}{2} \left( t-T \right)^2[/itex].

    OK, so then you say they collide when [itex]y_1=y_2[/itex]. That's good, but what does it tell you?
     
  7. Oct 10, 2012 #6
    oh yeah well its going upwards so yeah. well that the y's are the same. thats all i am able to assume
     
  8. Oct 10, 2012 #7
    My point is the equaton [itex]y_1=y_2[/itex] does not tell you at which height they collide. It does, however, provide some other useful information.
     
  9. Oct 10, 2012 #8
    such as the time it takes for them to collide? and their initial velocity perhaps?
     
  10. Oct 10, 2012 #9
    One of them, certainly. The initial velocity is already known though, as it is given in the problem description.
     
  11. Oct 10, 2012 #10
    i got that T=(2gt+1)/t
     
  12. Oct 11, 2012 #11

    Simon Bridge

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    Have you mixed up the roles of T and t?
    (Do you remember which time, t or T, you need?)
    Have you cancelled too many terms?
     
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