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Homework Help: Projectile motion of a thrown object

  1. Jun 29, 2011 #1
    1. an object is thrown upwardly from the roof of a building of height 30.8m with an initial velocity 23.8m/s at an angle 30degree to the horizontal. find the horizontal distance from the foot of the building to the point where the object hits the ground. assume g=9.8m/s^2
    answer is 82.4m

    2. i dont know what the question talking about especially this sentence "find the horizontal distance from the foot of the building to the point where the object hits the ground".
    give me some advice...thank you

    3. The attempt at a solution
  2. jcsd
  3. Jun 29, 2011 #2
    It is asking how far the object lands from the building. You know it lands 30.8 m below you, but you also know that tossing it forward means it is not immediately below the point where you let go.

    My students are usually helped a lot by a sketch. Draw the building and the path the object takes as it fall.
  4. Jun 29, 2011 #3


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    hi stupif! :smile:
    if the foot of the building is at (x,y) = (0,0),

    then the object starts at (0,30.8), and lands at (x,0), and you need to prove that x = 82.4 :wink:
  5. Jul 1, 2011 #4
    my solution

    1st , i find the the maximun height achieved.
    v^2 =u^2 +2as , 0= (23.8m/s sin 30)^2 +2(9.8)s
    s= 7.225m
    2nd, i find the time taken for achieved maximun height
    v=u+at, 0= 23.8 sin 30 + 9.8t
    t = 1.214
    3rd, i find the horizontal distance until the distance where the maximun height achieved.
    s= vt, s = 23.8cos 30 X 1.214s
    s= 25m
    4th, i find the horizontal distance from the distance where the maximun height achieved to the distance where the object hits the ground. (assume the initial angle is same as the final angle where the object reached the ground.
    tan 30 = (30.8 + 7.225)/ s
    s = 38/tan 30
    s = 65.82m

    total horizontal distance= 65.82 + 25=90.81m
    but the correct answer is 82.4m

    hope you all can understand what i am doing. thank you
  6. Jul 1, 2011 #5
    Your first three steps are good. Your assumption about the body landing at 30º, though, is not sound. That would mean moved in an arc until it reached its initial height and then went in a straight line. I suspect that is at odds with what you've casually noticed about projectiles.

    Do you know that projectiles move along a parabolic path? If so, symmetry tells you that the projectile returns to the height of 30.8 at x= 50 m and t= 2.429 s. Symmetry also tells you the velocity at that point is 23.8 m/s directed 30º below the horizontal. Do you see how to get the horizontal distance for the last leg of the trip from that information?

    (There are, I think, ways to do this all in fewer steps, but I recommend you continue on the path you started. If you like. I'll sketch them out for your later).
  7. Jul 1, 2011 #6
    i know is parabolic path.
    i dont really understand. what is the x=50? and t= 2.429s? how to get these?
    and symmetry means that the middle of the graph,then the middle point is the maximun height, then why the velocity is 23.8m/s ?why not 0m/s?
  8. Jul 1, 2011 #7
    Sorry: that would have been clear if we were both standing at a blackboard.

    Visualize the path of the projectile. It rises to the peak (which you showed is at x=25 m and t=1.214 s). When it returned to the original height (of 30.8 m) went another 25 m in another 1.214 s, so it is at x = 2(25 m ) = 50 m and t = 2(1.214 s) = 2.428 s.

    You are correct that the vertical component of the velocity is 0 m/s at the peak. The horizontal component remains a constant 23.8*cos(30) = 20.61 m/s since no forces act in the horizontal direction.

    At x=50 m, the vertical component of velocity has decreased to be -23.8*sin(30) = -11.9 m/s. You can see that from the symmetry, or you could calculate that dropping from vyi = 0 m/s at 9.8 m/s2 gives it that downward velocity after 1.214 s.

    Is that all making sense?

    If so, can you find how much longer it takes the object to land and how much farther it moves in that time?
  9. Jul 1, 2011 #8
    Hi all!
    An much simpler way to find the time taken for the object to lane is just using
    s = ut + 0.5at^2
    (-30.8) = (23.8 sin 30')t + 0.5(-9.8)t^2
    4.9t^2 - 11.9t - 30.8 = 0
    Solving the quadratic equation we can get the time travel from starting point until it lands on the ground, then just using
    s = vt
    where v = 23.8 cos 30'
  10. Jul 1, 2011 #9
    I agree it is simpler, but I thought it better to let stupif follow the path that first leapt out at him. Your approach is the one I referred to parenthetically in post #5

    For future reference, https://www.physicsforums.com/showthread.php?t=414380" ask that we not simply give answers to homework problems.
    Last edited by a moderator: Apr 26, 2017
  11. Jul 2, 2011 #10

    i understand 1st and 2nd paragraph, but in 3rd paragraph, -23.8sin 30 means what? and what you want me to do? find what?
  12. Jul 2, 2011 #11
    the answer become 25m again. t= 1.214 from quadratic equation, s = 25m ..
  13. Jul 2, 2011 #12
    Sorry, I'm new here.
    I'll avoid repeating the mistake. :blushing:
    Last edited by a moderator: Apr 26, 2017
  14. Jul 2, 2011 #13
    Sorry: I've been away from the keyboard all day.

    The vertical component of the velocity at the point the object returns to its original height is
    Vx=-23.8*sin(30) m/s.

    OK. Let's try this with a visual (the sketch below).

    Your intuition was to break path of the object into into parts. You found that it takes the projectile 1.214 s to reach point A and that point A is 25.0 m from the launch point.

    I then argued that, by symmetry, point B is another 25.0 m over and it takes another 1.214 s to get there. So the horizontal distance from the launch to B is 50 m.

    If you don't trust the symmetry argument, you can (and should) calculate the time to fall vertically from the height of A to the height of B given for that trip
    Viy=0 m/s
    a=-9.8 m/s2
    s = 7.225 m (which you calculated and posted in post #4)

    So, you've got how far the object went to reach B and you need to figure out how far it travels to get to C. You can rely on symmetry again to tell you that at B the horizontal velocity is Vx=23.8*cos(30) = 20.61 m/s and the vertical velocity is Vy=-23.8*sin(30) = -11.9 m/s. The initial height is 30.8 m and the final height is 0 m.

    Can you find the horizontal distance covered between B and C from that?

    Attached Files:

  15. Jul 2, 2011 #14
    Well, let me offer a belated welcome to Physics Forums. If you haven't read the posting guidelines, do so. I've run afoul of them a few times myself. They are a little daunting at first but are very effective at keeping discussions civil, efficient and productive.
  16. Jul 4, 2011 #15
    how come straight away get 82.4m...if i use s= vt, s= 23.8cos30 X 4s
    = 82.4m
    you tell me find the distance from B to C,
    1st, i use this formula, s= ut +1/2at^2 to find t
    30.8= -11.9t + 4.9t^2
    t= 4s and t= -1.57s

    2nd , i find s= vt
    s= 23.8cos 30 X 4
    = 82.4m
  17. Jul 4, 2011 #16
    You have two sign errors. We never said it explicit, but we have been assuming that up is positive and down is negative. That means g = -9.8m/s2. Then the second term on the right should be -4.9t2.

    The other error is subtler. We are implicitly taking the starting point as t= 0 s and s= 0 m. That means that the base of the building is at sy=-30.8 m. So you equation should be

    -30.8= -11.9t - 4.9t^2

    Giving t= -4 s and t= 1.57 s

    Verify that you can get 82.4 m with that. I REALLY encourage you see that MalcolmKee's first post makes sense to you, as its the way to do all this far more directly. If it doesn't make complete sense, ask and we'll get you to where you can use that approach next time.
  18. Jul 5, 2011 #17
    why gravity should be -9.8m/s^2, the object is toward gravity , right?
  19. Jul 5, 2011 #18
    Acceleration, velocity, force, displacement and other quantities are vectors, which means they have a size and and a direction. Direction can be given as north, southwest, etc, as an angle, as left or right... When the vector is being used in a calculation, we use + and - to show direction. I really should have shown on the diagram I posted that up and left are positive and down and right are negative. I picked those directions becasue of what you were calling positive in your earlier calculations.

    You can make the positive and negative directions go the opposite way if you want. You just need to be consistent. So when the object moves upward in the diagram as it is, ∆s and v are positive. When the object moves leftward, ∆s and v are positive. Since the acceleration is always downward, it is always negative.
  20. Jul 5, 2011 #19
    okay...thank you very much....
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