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    (Original post by morgan8002)
    No. Its original vertical velocity is independent of its horizontal velocity and the time taken to fall to the ground only depends on the original vertical velocity and the original height(if g is fixed).

    Hitting the ground at time t is the same constraint as the vertical displacement being 0(measured from the ground) at time t. The vertical displacement is just a function of the original vertical velocity and t, so if we set it equal to 0 and use the value of the original vertical velocity to work out t we have the time to hit the ground.
    This is the reason my teacher gave when i asked him about this problem but since i can't ask him i'll ask here on TSR

    Why is it independent? why doesn't adding things up work? surely the resultant of everything is what you have in that line the ball travels so surely splitting everything up into components shouldn't matter if you want to add them back up again right?

    (Original post by Eimmanuel)
    As a few already mention that the vertical velocity does not affect the horizontal velocity during the travel of the ball, but where the ball is located at each time, is influenced by the vertical velocity and horizontal velocity.

    Allow me to take a different approach, assume the initial horizontal velocity is 5 m/s. Take the height of the tower to be 50 m and the acceleration due to gravity to be 10 m/s^2.

    Initial coordinate is (0, 50)
    At t = 1 s, the ball is at coordinate (5, 50 - 0.5*10*1^2) = (5, 45)
    At t = 2 s, the ball is at coordinate (10, 50 - 0.5*10*2^2) = (10, 30)
    At t = 3 s, the ball is at coordinate (15, 50 - 0.5*10*3^2) = (15, 5)

    Next, consider a ball drop from the same tower from rest.
    Initial coordinate is (0, 50)
    At t = 1 s, the ball is at coordinate (0, 50 - 0.5*10*1^2) = (0, 45)
    At t = 2 s, the ball is at coordinate (0, 50 - 0.5*10*2^2) = (0, 30)
    At t = 3 s, the ball is at coordinate (0, 50 - 0.5*10*3^2) = (0, 5)

    In both cases, the ball falls 45 m in 3 seconds in regardless of the horizontal velocity.

    Hope it helps.
    .... he showed me a quick little experiment when he put a coin on the edge of a table and the ruler was over hanging the table and had another coin on the ruler

    what happened was when he moved the ruler quickly the coin on the edge of the table shot off in a parabola and the coin on the ruler dropped vertically down and they both hit the ground at the same time...
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    (Original post by thefatone)
    This is the reason my teacher gave when i asked him about this problem but since i can't ask him i'll ask here on TSR

    Why is it independent? why doesn't adding things up work? surely the resultant of everything is what you have in that line the ball travels so surely splitting everything up into components shouldn't matter if you want to add them back up again right?
    It's independent because if you increase the horizontal velocity it doesn't change the vertical velocity. Hitting the ground is the same as s=-h, where h is the initial height. s = ut + \frac{1}{2}at^2, so letting v_v be the initial vertical velocity and resolving vertically, -h = v_v t  - gt^2\Rightarrow gt^2 - v_v t - h = 0 \Rightarrow t = \dfrac{v_v \pm\sqrt{v_v^2 + 4gh}}{2g}
    \Rightarrow t = \dfrac{v_v + \sqrt{v_v^2 + 4gh}}{2g}\Rightarrow the time taken to hit the floor doesn't depend on the initial horizontal velocity.
    For the rest of it I'm not really sure what you mean.
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    (Original post by morgan8002)
    It's independent because if you increase the horizontal velocity it doesn't change the vertical velocity. Hitting the ground is the same as s=-h, where h is the initial height. s = ut + \frac{1}{2}at^2, so letting v_v be the initial vertical velocity and resolving vertically, -h = v_v t  - gt^2\Rightarrow gt^2 - v_v t - h = 0 \Rightarrow t = \dfrac{v_v \pm\sqrt{v_v^2 + 4gh}}{2g}
    \Rightarrow t = \dfrac{v_v + \sqrt{v_v^2 + 4gh}}{2g}\Rightarrow the time taken to hit the floor doesn't depend on the initial horizontal velocity.
    For the rest of it I'm not really sure what you mean.
    Ah ok right then.

    what i'm trying to say is vertical components + vertical components = resultant(which is the parabola the ball travels in
    so surely anything i do such as calculating time i would need to add up horizontal and vertical components to get the total time for example
 
 
 
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