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    A rubber bouncy ball bounces vertically into the air from ground level. You note that when it is 1.5 m above the ground it is travelling at 5 m s−1 upwards. Use conservation of energy to find (a) the ball’s speed just as it left the ground and (b) its maximum height, assuming no air resistance.

    How would I go about answer the above?
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    The sum of potential and kinetic energy is always the same. Therefore, E_k + E_p at 1.5 meters = E_k + E_p at 0 meters.

    a)
    So to find initial velocity, or v at 0 meters
    (E= 1/2mv^2+ mgh)
    E = 1/2 m(5)^2 + mg(1.5)
    E = 12.5m 1.5mg
    (Using E_k = 1/2mv^2 to solve for v considering mgh = 0 at 0 meters)

    v = 7.4 ms^-1 to 2sf

    b)
    At max height, the velocity of the ball is 0, therefore the energy is just equal to mgh
    (solving for h)
    12.5m 1.5mg = mgh
    h = 2.8m
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    (Original post by CaptainPegleg)
    The sum of potential and kinetic energy is always the same. Therefore, E_k + E_p at 1.5 meters = E_k + E_p at 0 meters.

    a)
    So to find initial velocity, or v at 0 meters
    (E= 1/2mv^2+ mgh)
    E = 1/2 m(5)^2 + mg(1.5)
    E = 12.5m 1.5mg
    (Using E_k = 1/2mv^2 to solve for v considering mgh = 0 at 0 meters)

    v = 7.4 ms^-1 to 2sf

    b)
    At max height, the velocity of the ball is 0, therefore the energy is just equal to mgh
    (solving for h)
    12.5m 1.5mg = mgh
    h = 2.8m
    Thank you, very helpful. (btw I'm currently in the 1st year of a natural sciences degree and I haven't done physics A-level so I'm a little bit behind)
 
 
 
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