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    hi please could someone help me, im doin maths alevel early, (currently in year10)

    Question: M1- Azmat throws a ball to suzan who is 80m away and who catches it at the same height as it was thrown. the ball is in the air for 5 seconds. Find the initial speed of the ball and the angle at which it was thrown.

    ive nearly finished the book (3 last exercise), and want to finish it by friday. thanks
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    what have you done so far?
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    Well you'll need the resolutes of the velocity first. Horizontal should be incredibly easy to find and you can use the uniform acceleration formulas for the vertical.
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    s=80, t=5, a=-9.8(in y) so by v=u+at v=0 at max point in y,
    0= Usin(theta) + (-9.8x2.5) ---- Usin(theta)=24.5
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    First work out the horizontal speed of the ball using v=s/t
    Next you know it takes 2.5s for the initial upwards velocity to reach 0. Hence use a=(v-u)/t to work out the initial vertical speed of the ball. Finally use trig to work out the initial speed and angle from the component velocities.
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    (Original post by roar558)
    First work out the horizontal speed of the ball using v=s/t
    Next you know it takes 2.5s for the initial upwards velocity to reach 0. Hence use a=(v-u)/t to work out the initial vertical speed of the ball. Finally use trig to work out the initial speed and angle from the component velocities.
    in v=s/t, i only have one value so there are 2 unknowns, so how can i do that?
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    (Original post by hassantahir)
    in v=s/t, i only have one value so there are 2 unknowns, so how can i do that?
    you have two values s ie the distance is 80m, t the time that the ball is in the air is 5s
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    (Original post by hassantahir)
    in v=s/t, i only have one value so there are 2 unknowns, so how can i do that?
    The time is 5 seconds and the distance is 80m.
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    (Original post by StephenP91)
    The time is 5 seconds and the distance is 80m.
    oh sorry i thought you meant something else by 's'
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    (Original post by hassantahir)
    oh sorry i thought you meant something else by 's'
    ? isn't s the accepted notation for distance?
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    (Original post by roar558)
    ? isn't s the accepted notation for distance?
    's' is displacement, 'd' is distance.
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    its my mistake, i hope i havent caused any confusion
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    (Original post by TheGrandmaster)
    's' is displacement, 'd' is distance.
    my mistake, but in this instance they are referring to the same value which is what we were discussing.
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    thanks roar558, got it right, thanks, its because i dont really take into mind the basic formulas like v=s/t. i only keep the 5 acceleration ones in mind. Thanks anyways
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    (Original post by hassantahir)
    thanks roar558, got it right, thanks, its because i dont really take into mind the basic formulas like v=s/t. i only keep the 5 acceleration ones in mind. Thanks anyways
    Makes little difference. You can still use them on that question.

     s  = \frac{(u + v)t}{2}

    Where:

    

s = s

u = v

v = v

a = 0

t = t

    You'll see that it just falls into  v = \frac{s}{t}
 
 
 
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