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  1. #1
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    Question velocity

    hey i really need urgent help and i know this is a maths forum but i think this forum is really good. Its my physics homework on suvat equations and maths is closely associated with physics

    I really dont get this question

    1.A particle falls freely from a rest. In the last second of its motion, it falls 5/9 of its total fall. How far did it fall?

    and

    2. A ball is dropped from a height of 20m and rebounds with a velocity which is 3/4 of the verlocity with which it hits the ground. What is the time interval between the first and second bounces?

    im really stuck, thanks ellie x
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  2. #2
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    Quote Originally Posted by mitchoboy View Post
    hey i really need urgent help and i know this is a maths forum but i think this forum is really good. Its my physics homework on suvat equations and maths is closely associated with physics

    I really dont get this question

    1.A particle falls freely from a rest. In the last second of its motion, it falls 5/9 of its total fall. How far did it fall?

    [snip]
    Let x1 be the total distance particle falls. Let t1 be the time taken to fall this distance. Then:

    u = 0
    a = g
    t = t1
    s = x

    Substitute all this into an appropriate straight line motion formula to get equation 1.


    If it falls 5/9 of total fall in last second, then it falls 4/9 of total fall in t1 - 1 seconds:

    u = 0
    a = g
    t = t1 - 1
    s = (4/9) x

    Substitute all this into an appropriate straight line motion formula to get equation 2.


    Solve equations (1) and (2) simultaneously for x and t1.
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  3. #3
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    Quote Originally Posted by mitchoboy View Post
    [snip]
    2. A ball is dropped from a height of 20m and rebounds with a velocity which is 3/4 of the verlocity with which it hits the ground. What is the time interval between the first and second bounces?

    im really stuck, thanks ellie x
    Calculate the velocity at which the ball hits the ground:

    u = 0
    s = 20
    a = g
    v = ?

    Use 3/4 of this v as the initial velocity for the upward motion after 1st bounce. Call this velocity u. Then:

    u = u.
    a = g.
    v = -u (assuming no air resistance)
    t = ?

    Note that v = -u because:

    1. Final speed at impact = initial speed
    2. Final speed is down, initial speed is up. Hence the sign change.

    Note that after second bounce, the speed is 3/4 of u.
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