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Motion in a vertical plan question

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    a stone is dropped from the top of a tower. 3 seconds later another stone is thrown downwards from the same point at 35m/s. if they hit the ground at the same time find the height of the tower.


    This is technically a mechanics 1 question but im sure the physics forum will have no trouble
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    S = ?
    u = 35 ms^-1
    v =
    a = 9.8 ms^-2 (because the particle is being dropped downwards we take gravity to be positive)
    t = 3s

    Now, use an equation that coincides with all of our measurements: in order to find s (the displacement).
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    (Original post by TarotOfMagic)
    S = ?
    u = 35 ms^-1
    v =
    a = 9.8 ms^-2 (because the particle is being dropped downwards we take gravity to be positive)
    t = 3s

    Now, use an equation that coincides with all of our measurements: in order to find s (the displacement).
    I already tried that. Like this:

    S = ut + 1/2at^2
    S = 35 (3) + 4.9(3^2)
    S = 105 + 44.1
    S = 149.5 m

    But according to the answer this is wrong it should be 579.5 m

    Thanks for your input though!
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    (Original post by hals)
    a stone is dropped from the top of a tower. 3 seconds later another stone is thrown downwards from the same point at 35m/s. if they hit the ground at the same time find the height of the tower.


    This is technically a mechanics 1 question but im sure the physics forum will have no trouble
    the struggle is real
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    (Original post by u.sha)
    the struggle is real
    Oh look we have matching avatars 😁
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    (Original post by TarotOfMagic)
    S = ?
    u = 35 ms^-1
    v =
    a = 9.8 ms^-2 (because the particle is being dropped downwards we take gravity to be positive)
    t = 3s

    Now, use an equation that coincides with all of our measurements: in order to find s (the displacement).
    You're an idiot. The stone moving at 35 m/s is dropped 3 seconds after the first, it doesn't take 3 seconds to reach the floor.
    (Original post by hals)
    a stone is dropped from the top of a tower. 3 seconds later another stone is thrown downwards from the same point at 35 m/s. if they hit the ground at the same time find the height of the tower.


    This is technically a mechanics 1 question but im sure the physics forum will have no trouble
    For the first stone,
    s=
    u=0
    v=
    a=9.8
    t=t+3

    For the second stone
    s=
    u=35
    v
    a=9.8
    t=t

    Equate the two displacements to find time, then find displacement using that value of t
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    (Original post by an_atheist)
    You're an idiot. The stone moving at 35 m/s is dropped 3 seconds after the first, it doesn't take 3 seconds to reach the floor.

    For the first stone,
    s=
    u=0
    v=
    a=9.8
    t=t+3

    For the second stone
    s=
    u=35
    v
    a=9.8
    t=t

    Equate the two displacements to find time, then find displacement using that value of t
    Thanks
    Shouldn't t=t-3 for the 1st stone because it happens before the 2nd one -.-


    P.s. That idiot remark was unnecessary but ill let it slide because you tried to help
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    (Original post by hals)
    Thanks
    Shouldn't t=t-3 for the 1st stone because it happens before the 2nd one -.-


    P.s. That idiot remark was unnecessary but ill let it slide because you tried to help
    **** you are correct. Sorry about the idiot thing, it annoys me when people (inclding myself) manage to mess up easy maths based problems
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    (Original post by an_atheist)
    **** you are correct. Sorry about the idiot thing, it annoys me when people (inclding myself) manage to mess up easy maths based problems
    Apology accepted lol. Obviously that 1st method i tried wouldnt have worked because t=3 was not the actual time of fall but then i didnt know wht else to do and came on here after trying that invalid method
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    (Original post by hals)
    Apology accepted lol. Obviously that 1st method i tried wouldnt have worked because t=3 was not the actual time of fall but then i didnt know wht else to do and came on here after trying that invalid method
    Nope, it is t+3 for the first stone. IF the second stone is falling for a time t, the first stone was released 3 seconds earlier, and was falling for t+3 seconds

    Got 579.5m for the height of the tower.
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    (Original post by an_atheist)
    **** you are correct. Sorry about the idiot thing, it annoys me when people (inclding myself) manage to mess up easy maths based problems
    Irony, at its finest.
 
 
 
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