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    can't do this dam question although im sure its not too difficult.

    tennis ball hit so that it moves vertically downwards from a height of 1m with an initial speed of 5ms^-1. when it hits the ground it rebounds vertically with half the speed it had when it hit the ground.

    Q - find the height to which it rebounds
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    (Original post by ShOcKzZ)
    can't do this dam question although im sure its not too difficult.

    tennis ball hit so that it moves vertically downwards from a height of 1m with an initial speed of 5ms^-1. when it hits the ground it rebounds vertically with half the speed it had when it hit the ground.

    Q - find the height to which it rebounds
    v^2 = u^2 + 2as

    v = (u^2 + 2gs)^1/2

    v = (25 + 2 x 9.8 x 1)^1/2

    Half that answer then use that as your initial speed and v = 0.

    v^2 = u^2 + 2as

    => s = (v^2 - u^2)/2g

    Remember to use g = -9.81.
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    Just using conservation of energy (taking g=9.81):

    mg1+m5²/2 = mv²/2 => v = 6.68 m/s when it hits the ground

    m(3.34)²/2 = mgh => h = 0.57m
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    (Original post by elpaw)
    Just using conservation of energy (taking g=9.81):

    mg1+m5²/2 = mv²/2 => v = 6.68 m/s when it hits the ground

    m(3.34)²/2 = mgh => h = 0.57m
    correct answer but i don't understand any of it
    i'll rep you tmrw
    thanks for helping.
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    (Original post by ShOcKzZ)
    correct answer but i don't understand any of it
    i'll rep you tmrw
    thanks for helping.
    dont rep me, rep nylex, he's got a correct method too. i believe my method is M3, so you wont get it just yet (if you havent done M3)
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    (Original post by elpaw)
    dont rep me, rep nylex, he's got a correct method too. i believe my method is M3, so you wont get it just yet.
    i did rep nylex
    but i will rep you for takin the time to reply..
    and im doing M3, but i its AQA B so its like M1 really..if that makes sense.
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    (Original post by ShOcKzZ)
    i did rep nylex
    but i will rep you for takin the time to reply..
    and im doing M3, but i its AQA B so its like M1 really..if that makes sense.
    oh ok, well its just conservation of energy, where:

    (GPE + KE)start = (GPE + KE)end
    where GPE = mgh and KE = mv²/2

    so for the first bit:

    mg1 + m5²/2 = mg0 + mv²/2
    m's cancel, so you get v²/2 = 9.81 + 12.5 = 22.36 => v²=44.72 => v = 6.68 m/s

    its speed is halved upon impact, so v = 3.34 m/s

    using CoE again:

    mg0 + m(3.34)²/2 = mgh + m0²/2
    m's cancel, so you get 5.5775 = 9.81h => h = 5.5775/9.81 = 0.57 m
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    Lol, thanx for rep (both of you) .
 
 
 
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