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Physics Constant Acceleration Question Watch

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    I have this question:

    A ball is thrown vertically upwards with an initial velocity of 14ms-1. Calculate:
    a) Maximum height reached
    b) Time taken to reach this height
    c) total time until ball is caught again
    d) what assumptions have to be made about (a) and (c)

    This is what I did:
    a)

    S
    U 14
    V 0
    A 9.81
    T

    v^2 = u^2 + 2as
    0 = 14^2 + 2(9.81)s
    -14^2 / 2(9.81) = s = 9.98 m

    b)
    s = 0.5(u+v)t
    9.98 = 0.5(14 + 0)t
    9.98 = 7t
    t = 1.43 sec (3sf)

    c) 1.43 * 2 = 2.86 sec
    d) Air resistance is negligible.

    Is this correct?
    Thanks, nwmyname.
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    (Original post by nwmyname)
    I have this question:

    A ball is thrown vertically upwards with an initial velocity of 14ms-1. Calculate:
    a) Maximum height reached
    b) Time taken to reach this height
    c) total time until ball is caught again
    d) what assumptions have to be made about (a) and (c)

    This is what I did:
    a)

    S
    U 14
    V 0
    A 9.81
    T

    v^2 = u^2 + 2as
    0 = 14^2 + 2(9.81)s
    -14^2 / 2(9.81) = s = 9.98 m

    b)
    s = 0.5(u+v)t
    9.98 = 0.5(14 + 0)t
    9.98 = 7t
    t = 1.43 sec (3sf)

    c) 1.43 * 2 = 2.86 sec
    d) Air resistance is negligible.

    Is this correct?
    Thanks, nwmyname.
    Mainly. Firstly is this a physics mechanics question or an m1 question, as physics uses g as 9.81 maths as 9.8 (weird i know). Either way assuming it should be 9.81 you've rounded wromg but it's otherwise correct (9.99 is the correct rounding). I suspect for the last question they're probably looking for something more specific to the situation such as that the height the ball is released from is the same height that you would catch the ball at but its difficult to tell
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    In terms of a.) it is possible to calculate the maximum height of the ball by the velocity at the beginning and gravitation. It is

    h(max) = (v0)^2/2g = (14 ms^-1)^2 / 2 * 9,81 m/s^2 = 9,99 (rounded). Your result is a bit wrong due to rounded result.

    put that result for the height in path-time-law and convert to time t you get for b.)

    s = g/2* t^2 => t = square root 2s/g = square root 2*9,98/9,81 = 1,43 seconds (rounded). This result is right too.

    WHat did you mean when you are talking about 'caught' in c.)? when the ball has arrived the ground? or when the ball begins to fall to the ground?

    What you said in d.) is right. All the caluclations for vertical throws based on a negligible air resistance (and falling motions in a vertical way).
 
 
 
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