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Kinematics: Initial velocity of second rock if 2 rocks are thrown up?

  1. Jan 25, 2012 #1
    The problem statement, all variables and given/known data
    Rock A is thrown straight up from the top of a 25m tall building at a speed of 20m/s. Exactly 1 second later Rock B is thrown in such a way that it lands on the ground at the base of the building at the same instant that Rock A lands. What is the initial velocity for Rock B? (Air friction is neglected)

    Answer: 14m/s [up]

    ΔtA - 1 = ΔtB
    ViA = 20m/s [up]
    a = 9.8m/s^2 [down]
    Δd(building) = 25m
    ViB = ?

    My attempt:

    Rock A going up: vf^2 = vi^2 + 2aΔd
    0^2 - 20^2 = 2(-9.8)Δd
    Δd = 20.408m

    t = (2Δd)/(vf + vi)
    = (2 x 20.408)/20
    t = 2.04s

    Rock A going down: Δd = 20.408 + 25 = 45.408m

    Δd = viΔt + 0.5a(Δt)^2
    45.408 = 0.5 (-9.8)(Δt)^2
    Δt = 3.04s

    Rock A going up and then down: Δt = 2.04 + 3.04 = 5.08s

    Rock B going up and then down: 4.08s (5.08 - 1 because of the 1s delay in throwing)

    Then from here I'm stuck. I know the total time Rock B takes to go up and down is 4.08s. I know the building is 25m tall. I know acceleration is 9.8m/s^2 [down]. However, I don't know how those 4.08s are split between going up and down, nor do I know the distance that Rock B travels up before falling down. I already tried calculating Rock B's initial velocity if it was dropped instead of thrown (question doesn't specify, only "thrown in such a way"), but the answer isn't right.

    I also attempted to substitute.
    Rock B going up: t1 = Δv/a = vi/a
    Rock B going down: t2 = √(vi^2 + 50 x 9.8 / 9.8^2)
    t1 + t2 = 4.08s and substitute the above to solve for vi, but that didn't come out to the right answer either.

    Any hints or help on how to work with what I have would be greatly appreciated.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jan 25, 2012 #2
    Why have you started off with Δd = 25m then later on it's become Δd = 20.408m?
    You also seem to have assumed that the final velocity is zero, this will not be the case.

    You seem to be misunderstanding what you're trying to do here, I'll try and get you going in the right direction.

    First, I'd drop the whole [up] and [down] notation, it seems handy and intuitive now but later on down the line it'll come back to bite you. Pick a direction (most people will chose up but it's really up to you) and set this as your positive direction, then anything gonig down negative.

    So the boundary conditions for your problem are;

    [itex]y_{A}(0) = y_{B}(0) = 25 m[/itex]

    [itex]y_{A}(t_{final}) = y_{B}(t_{final}) = 0[/itex]

    [itex]v_{A}(0) = 20 \frac{m}{s}[/itex]

    [itex]v_{B}(0) = ?[/itex]

    and of course you have [itex] g = -9.8 \frac{m}{s^2}[/itex]

    Here im writing the y position as a function of time, y=y(t)
    I'm setting t=0 the instant the first rock is thrown.

    I don't know if you know any calculus, I'll assume you don't for now;
    You should know the equation of motion

    [itex] \Delta y = v_{initial} t + \frac{a}{2} t^2[/itex]

    Now, we should be careful to remeber what the change in y, [itex] \Delta y[/itex] is.
    The change is the position at the final time minus the position at the first time, which will give us -20
    We also substitute in the boundary conditions;

    [itex]y_{A}(t_{final}) - y_{A}(0) = v_{A}(0) t - \frac{g}{2} t^2[/itex]

    [itex]-20 = 20 t - \frac{9.8}{2} t^2[/itex]

    [itex]-\frac{9.8}{2} t^2 +20t +20=0[/itex]

    Which is just your standard quadratic equation which you can use to solve for the time the first rock lands.
    Hopefuly now you should understand what's going on here and you should be able to see where to go next.
  4. Jan 26, 2012 #3
    Thank you so much for taking the time to help me.

    The change in Δd that you mentioned happened because I split the rock's motion into being thrown straight up (which is the part where final velocity is zero), and falling down to the ground from its maximum height (final velocity is not zero here).

    Because the rock is being thrown straight up, I thought the Δd would not simply be 25m (the building height) because the height that it's going above the building needs to be added (which I calculated to be 20.408m).

    I took your advice and used this,

    [itex]-\frac{9.8}{2} t^2 +20t +25=0[/itex]

    substituting +25 for +20 because I'm assuming you meant for the change in y to be the height of the building.

    The positive value of t was about 5.08, which is what I arrived at in my calculations as well. The next step I took was to subtract a second to arrive at the time it takes for Rock B to land.

    So [itex]t_{A}=5.08s [/itex] and [itex]t_{B}=4.08s[/itex]

    I put 4.08 as t into the motion equation to solve for [itex]v_{initial}[/itex],

    [itex] -25 = v_{initial} (4.08) + \frac{-9.8}{2} (4.08)^2[/itex]

    using -25 as the [itex] \Delta y[/itex] like you did, and the [itex]v_{initial}[/itex] comes out to about 13.86, rounding to 14m/s - the right answer.

    So I should have just ignored the fact that the rock goes up before it comes down, and just focused on the displacement from starting point to ending point.

    Thanks again for your help!
  5. Jan 26, 2012 #4
    You're correct, I was a bit tired when I wrote it :p

    Yes, once you begin using calculus in physics you'll learn that what you're doing here is solving an ordinatry differential equation with boundary conditions.
    What happens in the middle doesn't really have any importance in these cases, as long as the solution satistifies the boundary conditions.
    The solution of the equation of motion in this problem is


    As you can see (I don't know if you know about deriviatives, if you don't you'll have to trust me on this) this satisfies the boundary conditions

    [itex]y(0)=25 + 20*0 -\frac{9.8*0}{2} = 25[/itex]
    [itex]y'(0)=v(0) = 20 - 9.8*0 = 20[/itex]
    [itex]y''(0)=a(0) = -9.8 [/itex]

    The first thing you did was to solve the equation y(t)=0 (the final position) for t which is where that equation that you're usually just told to remember comes from


    Which can be rearranged as

    [itex]0-25 = 20t - \frac{9.8}{2} t^2 [/itex]
    [itex]\Delta y = 20 t - \frac{9.8}{2} t^2[/itex]

    As you can see in our derivation nothing that happened in the middle came in to play so we don't need to concern ourselves with the fact that the rock went up and came abck down, it's the displacement from the starting point that matters here rather than the total distance traveled.
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