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Motion of two balls

  1. Dec 31, 2011 #1
    1. The problem statement, all variables and given/known data

    A ball is thrown vertically upwards at 5 m / s from a roof top of 100m. The ball B is thrown down from the same point 2s later at 20 m / s. Where and when will they meet.


    2. Relevant equations

    S = So + ut + 1/2at^2

    3. The attempt at a solution

    well at first I separated the fact that while ball B falls at say t seconds, ball A's time is given by t+2 (due to it being thrown 2 seconds prior to ball B).

    given what's given above (the velocity and height)...I wrote down the following equations to find TIME, where lower a and b denote ball A and B:

    Sa = Soa + ua(t+2) + 1/2a(t+2)^2
    Sb = Sob + ubt + 1/2at^2

    but i tried equalizing the above two equations with all the proper variables replaced by the values provided in the problem, but unfortunately i don't get the right answer.

    I'm lost, I'm not even sure if I tackled the problem correctly.
     
  2. jcsd
  3. Dec 31, 2011 #2

    SammyS

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    Science Advisor
    Homework Helper
    Gold Member

    Hello mechanics_boy. Welcome to PF !

    What answer do you get?

    What's the correct answer?

    What did you use for ua, ub, and a ?
     
  4. Dec 31, 2011 #3
    Ua = initial velocity of ball A
    Ub = initial velocity of ball B
    a = acceleration = g = -9.8 m/s^2

    Replacing the values known for the variables in the equation I get the following system:

    Ball A: Sa = 100 + 5(t+2) + 1/2(-9.8)(t+2)^2
    Ball B: Sb = 100 + 20t + 1/2(-9.8)t^2

    therefore the equation to solve:

    Sa = Sb
    100 + 5(t+2) + 1/2(-9.8)(t+2)^2 = 100 + 20t + 1/2(-9.8)t^2

    the answer according to the solution manual is t = 3,78s.

    I believe my whole way of looking at this problem is wrong. what do you think?
     
  5. Dec 31, 2011 #4

    Nugatory

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    Staff: Mentor

    Check all your signs carefully... Velocities and accelerations down are negative, up are positive (or vice versa, but you have to be consistent)
     
  6. Dec 31, 2011 #5
    Ok So Ub should be -20 m/s instead of 20/s because the ball B is thrown downwards. (thanks Nugatory)

    But adjusting the signs STILL doesn't give me the right answer.

    Am I completely wrong to even have established the above system of equations? Am I suppose to take this a completely different way..???
     
    Last edited: Dec 31, 2011
  7. Dec 31, 2011 #6

    jedishrfu

    Staff: Mentor

    since you have the answer why not use it check your eqns. if the vel and acc signs are correct then what is left is the time right? The 1st eqn is the simplest and its answer could be the clue.
     
  8. Dec 31, 2011 #7

    ehild

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    Gold Member

    There can be mistakes in an answer key. It happens quite often. If you replace ub with the correct value ub=-20 your equations are correct. What result did you got?

    ehild

    btw: Happy New Year!
     
  9. Jan 1, 2012 #8
    Ok, I got the right answer. Main mistake as pointed out was the sign for Ub, should be negative and once corrected I plugged into the equation and got t=1,78s. However this is the time taken for ball B to reach the same height as ball A. As for ball A, time taken for it to reach the same height as B is 2+1,78=3,78s as it has been thrown 2 seconds prior than ball B.

    given time, i found they joined each other at the height of 48,9m.

    thank you to all, I can't believe I was incredibly stupid with this number, hopefully this won't repeat again.

    thank you to: SammyS, Nugatory, Jedishrfu and ehild!!! :)
     
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