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    A ball thrown directly upwards rises to a highest point and falls back into the throwers hand; what is the average velocity?
    By \DeltaDistance=Average Velocity x \Deltatime
    the Va would be O since \Deltadistance is 0, which is what the answer says.

    But then if I was to go by Average velocity = (Final velocity + Initial velocity)/2
    wouldn't it not be zero since there had to be some final velocity on its way down?
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    (Original post by eterna)
    By \DeltaDistance=Average Velocity x \Deltatime
    the Va would be O since \Deltadistance is 0, which is what the answer says.

    But then if I was to go by Average velocity = (Final velocity + Initial velocity)/2
    wouldn't it not be zero since there had to be some final velocity on its way down?
    If the final velocity is +v then the initial velocity will be -v (opposite direction)
    Final + initial = (+v + (- v) )/2 = 0
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    (Original post by Stonebridge)
    If the final velocity is +v then the initial velocity will be -v (opposite direction)
    Final + initial = (+v + (- v) )/2 = 0
    thanks! Does the negative v for the opposite direction also apply when finding average velocity vf+vi/2 ?
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    (Original post by eterna)
    thanks! Does the negative v for the opposite direction also apply when finding average velocity vf+vi/2 ?
    Yes, but you need to be careful with this.
    I should have pointed out from your 1st post that it should be
    distance = average speed x time
    displacement = average velocity x time

    distance and speed are scalars
    displacement and velocity are vectors

    If you throw a ball up to a height of, say 2m, and catch it the total displacement is zero. The distance travelled is 4m.

    The average speed will be 4m / time taken
    The average velocity will be 0m / time taken


    If you travel round a circle of radius r, your displacement is 0. Your distance travelled is 2πr. Your average velocity and average speed will consequently be different.
    It's an important distinction in physics.

    Average speed is defined as total distance travelled divided by time taken.
    Average velocity is (v1 + v2)/2 only for uniform acceleration in a straight line. In general it's total displacement / time taken.
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    (Original post by Stonebridge)
    Yes, but you need to be careful with this.
    I should have pointed out from your 1st post that it should be
    distance = average speed x time
    displacement = average velocity x time

    distance and speed are scalars
    displacement and velocity are vectors

    If you throw a ball up to a height of, say 2m, and catch it the total displacement is zero. The distance travelled is 4m.

    The average speed will be 4m / time taken
    The average velocity will be 0m / time taken


    If you travel round a circle of radius r, your displacement is 0. Your distance travelled is 2πr. Your average velocity and average speed will consequently be different.
    It's an important distinction in physics.

    Average speed is defined as total distance travelled divided by time taken.
    Average velocity is (v1 + v2)/2 only for uniform acceleration in a straight line. In general it's total displacement / time taken.
    Thank you very much!
    One last question, given the current information, how should I proceed with finding x?

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    (Original post by eterna)
    Thank you very much!
    One last question, given the current information, how should I proceed with finding x?


    Could you post the complete question, please.
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    (Original post by Stonebridge)
    Could you post the complete question, please.
    Well, it wasn't a really a question; they just used values already known to calculate the range.

    I just wanted to know that given a question with a negative angle( I can't find a proper one), how I should proceed with finding the range.


    If the angle had been positive, I'd find the time it took for the object to reach its max vertical height and then multiply that by 2 to get the time it took for the object to reach the endpoint horizontally
    The initial velocity would have been given and I'd take the horizontal component of that.
    Id then use s=ut to get the range


    But with a negative angle I don't know how to proceed.
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    (Original post by eterna)
    Well, it wasn't a really a question; they just used values already known to calculate the range.

    I just wanted to know that given a question with a negative angle( I can't find a proper one), how I should proceed with finding the range.


    If the angle had been positive, I'd find the time it took for the object to reach its max vertical height and then multiply that by 2 to get the time it took for the object to reach the endpoint horizontally
    The initial velocity would have been given and I'd take the horizontal component of that.
    Id then use s=ut to get the range


    But with a negative angle I don't know how to proceed.
    The distance x is 25t where t is the time of flight (horizontal motion)
    The distance y = ½gt² (from s = ut + ½gt² on the vertical motion)

    You also know y/x = tan 35
    from this you can find t
    From t you can find x and y
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    (Original post by Stonebridge)
    The distance x is 25t where t is the time of flight (horizontal motion)
    The distance y = ½gt² (from s = ut + ½gt² on the vertical motion)

    You also know y/x = tan 35
    from this you can find t
    From t you can find x and y
    alright, thanks!
 
 
 
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