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Vertical Motion

  1. Apr 26, 2009 #1
    1. The problem statement, all variables and given/known data
    "Two boys stand on the roof of an office block. One drops a ball from the roof, the other throws a ball up vertically with an initial velocity of 15 ms^-1. The balls strike the ground 1.88s apart. Calculate the heigh of the office block"

    2. Relevant equations

    x = x_0 + v_0 t + (1/2) a t^2

    I learnt formulae with different symbols, and so I found it difficult to recognise relevant equations :(

    3. The attempt at a solution

    (no idea if ANY of this is correct so far)
    I said the ball going up was ball 1, and the ball which is dropped was ball 2.
    I drew a diagram, and worked with ball 1. Firstly, i calculated the time it took for ball 1 to reach a velocity of 0 m/s, which was 1.53 seconds. Then, I calculated the distance from the rooftop to this point, labelled P, which was 11.48 metres. I also labelled office block height as h, and then I said that ball 1 would fall from P to the ground a total of h + 11.48 metres. The time ball 1 takes from P to the ground would be the time that ball 2 takes, plus 1.53 plus 1.88.
    My wording was most likely incredibly confusing and incoherent, but basically I'm stuck now. I've tried forming equations such as h+11.48 = 1/2 9.8(t + 3.41)^2 but I suspect I'm forming incorrect equations and have terrible methods...can anybody please take me through this question? I know you won't DO the question which is good - i need to learn it myself - but could you please point out errors in my working and perhaps help me form a suitable equation to solve? :)
     
  2. jcsd
  3. Apr 26, 2009 #2

    Doc Al

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    Staff: Mentor

    You're thinking isn't too far off. To make it more systematic, start with this equation:
    For each case, dropped ball and thrown ball, write that equation with the appropriate parameters for the final position of the ball.

    What's x and x_0? (Hint: I suggest you call the ground 0 and the height of the building h.)
    What's v_0 in each case?
    What's a?
    Call the times t_1 and t_2. What's the equation connecting them?

    When you have your equations, you'll be able to solve for h. (Messy, but doable.)
     
  4. Apr 26, 2009 #3
    Sorry, I'm not exactly sure what x and x_0 are - the displacement formula I learnt was
    r = ut + 1/2 a t^2. where I guess u is v_0. Will work on the rest.

    v_0 is 15ms^-1 for thrown ball, and 0 ms^-1 for dropped ball.
    so a is -9.8 on the way up and 9.8 on the way down.

    what exactly is t_1? should it be the time from the moment the ball is thrown to the moment it touches the ground? or from when it reaches its peak, to the moment it touches the ground? t_2 = t_1 - 1.88?

    will add more if I get any further
     
    Last edited: Apr 26, 2009
  5. Apr 26, 2009 #4

    Doc Al

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    Staff: Mentor

    x is the final position at time t; x_0 is the initial position at time 0. (To relate it to what you already know, x - x_0 would be the displacement.)

    What are the initial (x_0) and final (x) positions of the ball in each case?

    Yes, v_0 is the initial velocity (what you learned as u). What is it in each case?
     
  6. Apr 26, 2009 #5

    Doc Al

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    Staff: Mentor

    Good.
    Careful! The acceleration is -9.8 m/s^2 at all times. (It doesn't change sign.)

    Yes.
    No, measure time from the starting point.

    Yes.
     
  7. Apr 26, 2009 #6
    How about this:
    the thrown ball takes 0.35 seconds to reach the ground from when it reaches the roof again, 1.88 - 1.53 if I say the ball is coming from the 11.48 m mark.
    After this, I form a pair of simultaneous equations to solve:

    h = 1/2 9.8 t^2
    = 4.9t^2 (1)
    h + 11.48 = 4.9 (t+0.35)^2 (2)
    substitute (1) into (2)
    I don't want to do the working here, but I got t = 3.17 seconds. sub t = 3.17 seconds back in to (1) and I got
    h = 4.9 x 3.17^2
    = 49.48 metres

    is this correct? I get confused easily so i probably mixed up some values. I know I definitely ignored one (or more?) of your tips because I thought I was having a revelation!
     
  8. Apr 27, 2009 #7

    Doc Al

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    Staff: Mentor

    Excellent work! :approve:

    There are many ways to solve these problems. Your method is just fine. (And it's almost always better to figure things out in your own way--as long as you're correct. :wink:)

    Just for the record, here's what I suggested:

    Using the standard formula (x = x_0 + v_0*t + .5at^2) I set up two equations:

    For the dropped ball: 0 = h -.5*9.8*(t_1)^2
    For the thrown ball: 0 = h +15*(t_2) -.5*9.8*(t_2)^2

    Substitute t_2 = t_1 + 1.88 in the second equation, solve them simultaneously, and you'll get the same answer as you did for h.

    (Sorry it took me so long to respond; got a bit busy yesterday.)
     
    Last edited: Apr 27, 2009
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