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    (Original post by thefatone)
    Help with 5 a pls i have no idea about waves >.> they're my worst
    Sorry I'm on my phone atm so can't use LaTeX. 5.a looks like you need to understand what a stationary wave is, it's basically when two progressive waves go in opposite directions, reflect and then interfere with each other at the same time. The waves have similar amplitudes
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    (Original post by The-Spartan)
    Sorry I'm on my phone atm so can't use LaTeX. 5.a looks like you need to understand what a stationary wave is, it's basically when two progressive waves go in opposite directions, reflect and then interfere with each other at the same time. The waves have similar amplitudes
    what do they reflect of off?
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    (Original post by thefatone)
    what do they reflect of off?
    There are 2 closed ends to the string, that the waves will reflect off I'll be like 20 mins before I get home, I'll try and draw a diagram to show it better

    Edit: make that 40 mins (i hate traffic -_-)

    anyway here is an animation showing it. It kind of looks like this:
    Spoiler:
    Show
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    (Original post by thefatone)
    ^^ you've perfectly demonstrated why i think the path where the ball is thrown is longer since you've just made a triangle, the path where the ball is thrown looks like the longest side, if the ball has more distance to travel the time taken cannot be the same.

    But why is horizontal velocity not counted? that's because we're doing suvat for vertical you can use  v=\dfrac{s}{t} for horizontal

    well ffs then -.-

    Lol thanks for backing me up.. dunno what's with that guy :/
    Oh noes :beard:
    The horizontal component does not contribute to the rate at which the ball falls, so therefore the ball hits the ground at the same time. Its when the ball hits the ground that the timer stops
    They will both travel different distances yes, but that is only done from the horizontal component.

    And no, if the ball if accelerating you cannot use v=\frac{s}{t} because the velocity is changing all the time. Thats what SUVAT is for :yes:
    you could use average velocity but thats still not as good as SUVAT.

    and yeah no worries. not sure either
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    (Original post by The-Spartan)
    Oh noes :beard:
    The horizontal component does not contribute to the rate at which the ball falls, so therefore the ball hits the ground at the same time. Its when the ball hits the ground that the timer stops
    They will both travel different distances yes, but that is only done from the horizontal component.

    And no, if the ball if accelerating you cannot use v=\frac{s}{t} because the velocity is changing all the time. Thats what SUVAT is for :yes:
    you could use average velocity but thats still not as good as SUVAT.

    and yeah no worries. not sure either

    If ball is accelerating but for horizontal it's not accelerating?
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    (Original post by thefatone)
    If ball is accelerating but for horizontal it's not accelerating?
    Aha silly me i thought you meant for the vertical . yes, the horizontal velocity is v=\frac{s}{t}. The time in this case though is the time taken for the ball to hit the ground no? this is solely dependant on the vertical acceleration/velocity. can you now see how the time here is how long it takes the ball to hit the ground, and thus the time to drop a ball and throw one is the same?
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    (Original post by The-Spartan)
    Aha silly me i thought you meant for the vertical . yes, the horizontal velocity is v=\frac{s}{t}. The time in this case though is the time taken for the ball to hit the ground no? this is solely dependant on the vertical acceleration/velocity. can you now see how the time here is how long it takes the ball to hit the ground, and thus the time to drop a ball and throw one is the same?
    hmmmmm so for the thrown one i'd need to calculate horizontally and vertically time taken then add them up and that would = time take for ball to drop vertically?
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    (Original post by thefatone)
    hmmmmm so for the thrown one i'd need to calculate horizontally and vertically time taken then add them up and that would = time take for ball to drop vertically?
    Not quite

    For the dropped one right:
    Time is basically from when you release the ball to it hitting the ground.
    So you can calculate how long it takes to hit the ground using the vertical component yeah?

    The thrown one:
    Time is still, as before, from when you throw it to it hitting the ground
    the only thing pulling the ball is the vertical component
    As such the horizontal has 0 effect on the time. Have a look here again
    Spoiler:
    Show
    Here you see the ball being dropped. It takes 4 seconds for the ball to hit the ground right, in 4 steps here.

    For throwing the ball:

    You see it only takes the ball again, 4 seconds to reach the same point in the y axis. This is the time taken for the ball to hit the ground. See how they both hit the ground at 4 seconds, even though the curved path was thrown? This does not matter how fast you throw it, it just makes the curve flatter thats all:

    Here is a ball thrown not as fast as previous. Again it hits the ground at 4 seconds.

    Time in this context is the time taken for the ball to hit the ground
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    (Original post by The-Spartan)
    Not quite

    For the dropped one right:
    Time is basically from when you release the ball to it hitting the ground.
    So you can calculate how long it takes to hit the ground using the vertical component yeah?

    The thrown one:
    Time is still, as before, from when you throw it to it hitting the ground
    the only thing pulling the ball is the vertical component
    As such the horizontal has 0 effect on the time. Have a look here again
    Spoiler:
    Show
    Here you see the ball being dropped. It takes 4 seconds for the ball to hit the ground right, in 4 steps here.

    For throwing the ball:

    You see it only takes the ball again, 4 seconds to reach the same point in the y axis. This is the time taken for the ball to hit the ground. See how they both hit the ground at 4 seconds, even though the curved path was thrown? This does not matter how fast you throw it, it just makes the curve flatter thats all:

    Here is a ball thrown not as fast as previous. Again it hits the ground at 4 seconds.

    Time in this context is the time taken for the ball to hit the ground
    If i'm powerful enough to throw that ball a mile and we know where that ball lands, i'm pretty sure if the was the thrown ball and dropped ball they wouldn't land on the ground at the same time, i'm assuming your telling me horizontal distance doesn't mateer
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    (Original post by thefatone)
    If i'm powerful enough to throw that ball a mile and we know where that ball lands, i'm pretty sure if the was the thrown ball and dropped ball they wouldn't land on the ground at the same time, i'm assuming your telling me horizontal distance doesn't mateer
    Nope as long as the ball is thrown in a straight line (no upward motion) they will land at the same time. Hard concept to get your head around but yes, horizontal distance has no effect on overall time

    think about it the ball you throw at a huge initial horizontal velocity would still have downwards acceleration of 9.81, so would be pulled to the earth at the same rate, and therefore will it the floor at the same time

    Dont be put off if you dont get it straight away, i had to contemplate it for a good solid 3 hours before i understood it. It's kind of counter intuitive!
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    (Original post by thefatone)
    If i'm powerful enough to throw that ball a mile and we know where that ball lands, i'm pretty sure if the was the thrown ball and dropped ball they wouldn't land on the ground at the same time, i'm assuming your telling me horizontal distance doesn't mateer
    Horizontal distance has no effect at all on the time it takes for the ball to hit the ground. Everything we are dealing with here are vector quantities. Throwing the ball horizontally (i.e. at an angle of 0degrees to the horizontal) will not have any vertical component (can be seen intuitively or because it would be vsintheta and theta is 0), so when we consider the vertical motion it is still just being accelerated at 9.81ms-2 downwards, regardless of how fast it is travelling horizontally
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    (Original post by The-Spartan)
    Nope as long as the ball is thrown in a straight line (no upward motion) they will land at the same time. Hard concept to get your head around but yes, horizontal distance has no effect on overall time

    think about it the ball you throw at a huge initial horizontal velocity would still have downwards acceleration of 9.81, so would be pulled to the earth at the same rate, and therefore will it the floor at the same time

    Dont be put off if you dont get it straight away, i had to contemplate it for a good solid 3 hours before i understood it. It's kind of counter intuitive!
    ..... i understand the words and reasons why but it doesn't agree with my logic
    (Original post by samb1234)
    Horizontal distance has no effect at all on the time it takes for the ball to hit the ground. Everything we are dealing with here are vector quantities. Throwing the ball horizontally (i.e. at an angle of 0degrees to the horizontal) will not have any vertical component (can be seen intuitively or because it would be vsintheta and theta is 0), so when we consider the vertical motion it is still just being accelerated at 9.81ms-2 downwards, regardless of how fast it is travelling horizontally
    i understand but my logic tells me otherwise.

    thanks both for trying to get me to understand but i can can't the doubt that if i trow a ball super hard and drop another ball at the same height as the thrown balls i can't see how the tame taken to get to the ground would ever be the same.
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    (Original post by thefatone)
    ..... i understand the words and reasons why but it doesn't agree with my logic


    i understand but my logic tells me otherwise.

    thanks both for trying to get me to understand but i can can't the doubt that if i trow a ball super hard and drop another ball at the same height as the thrown balls i can't see how the tame taken to get to the ground would ever be the same.
    Why wouldnt it be?
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    (Original post by thefatone)
    ..... i understand the words and reasons why but it doesn't agree with my logic


    i understand but my logic tells me otherwise.

    thanks both for trying to get me to understand but i can can't the doubt that if i trow a ball super hard and drop another ball at the same height as the thrown balls i can't see how the tame taken to get to the ground would ever be the same.
    Assuming you are throwing the ball horizontally and not upwards (0 degrees to the horizontal) then yes they will hit the ground at the same time.

    There are a few videos showing this:
    https://youtu.be/zMF4CD7i3hg?t=1m4s
    http://www.discovery.com/tv-shows/my...-fired-bullet/

    Your logic assumes that the ball keeps off the ground because of its horizontal velocity, whereas the ball hits the ground at the same time regardless of horizontal velocity
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    (Original post by samb1234)
    Why wouldnt it be?
    because it travels a horizontal distance too and example which shows my thinking is this

    a triangle with angles 45 45 and 90
    one of the sides is perpendicular to the other and i would say that would be the *distance the ball was dropped* and the hypotenuse of that triangle would be *the distance the ball is thrown* by this logic ball thrown should take more time since it's a greater distance.

    (Original post by The-Spartan)
    Assuming you are throwing the ball horizontally and not upwards (0 degrees to the horizontal) then yes they will hit the ground at the same time.

    There are a few videos showing this:
    https://youtu.be/zMF4CD7i3hg?t=1m4s
    http://www.discovery.com/tv-shows/my...-fired-bullet/

    Your logic assumes that the ball keeps off the ground because of its horizontal velocity, whereas the ball hits the ground at the same time regardless of horizontal velocity
    yea horizontal dude not up xD

    not the horizontal velocity but the extra distance it has to travel going horizontally
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    (Original post by thefatone)
    because it travels a horizontal distance too and example which shows my thinking is this

    a triangle with angles 45 45 and 90
    one of the sides is perpendicular to the other and i would say that would be the *distance the ball was dropped* and the hypotenuse of that triangle would be *the distance the ball is thrown* by this logic ball thrown should take more time since it's a greater distance.



    yea horizontal dude not up xD

    not the horizontal velocity but the extra distance it has to travel going horizontally
    Yes it does travel a longer distance, but it travels at a higher velocity. The vertical component is still the only thing that matters in regards to how long it takes to hit floor which isnt affected in any way
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    (Original post by thefatone)
    because it travels a horizontal distance too and example which shows my thinking is this

    a triangle with angles 45 45 and 90
    one of the sides is perpendicular to the other and i would say that would be the *distance the ball was dropped* and the hypotenuse of that triangle would be *the distance the ball is thrown* by this logic ball thrown should take more time since it's a greater distance.



    yea horizontal dude not up xD

    not the horizontal velocity but the extra distance it has to travel going horizontally
    It's travelling faster than a ball that is just dropped. It has the horizontal velocity as well, which cancels with the extra distance to give the same time.

    We're assuming that the Earth is flat and there's no air resistance for these calculations. If you add either of these then horizontal velocity changes things.
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    (Original post by samb1234)
    Yes it does travel a longer distance, but it travels at a higher velocity. The vertical component is still the only thing that matters in regards to how long it takes to hit floor which isnt affected in any way
    surely you need to take into account both time taken to travel horizontally and vertically to get the total time to reach the ground??
    (Original post by morgan8002)
    It's travelling faster than a ball that is just dropped. It has the horizontal velocity as well, which cancels with the extra distance to give the same time.

    We're assuming that the Earth is flat and there's no air resistance for these calculations. If you add either of these then horizontal velocity changes things.
    ^^ same reply as above about horizontal and vertical stuff
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    (Original post by thefatone)
    surely you need to take into account both time taken to travel horizontally and vertically to get the total time to reach the ground??
    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.
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    (Original post by thefatone)
    surely you need to take into account both time taken to travel horizontally and vertically to get the total time to reach the ground??


    ^^ same reply as above about horizontal and vertical stuff

    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.
 
 
 
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