Physics

A ball rolls horizontally off the edge of a tabletop that is 1.80 m high. It strikes the floor at a point 1.42 m horizontally away from the table edge. (Neglect air resistance.)
(a) How long was the ball in the air?
(b) What was its speed at the instant it left the table?

i have the equation
x-x(0)= V(0)xt + 1/2at^2
y-y(0) = V(0)yt - 1/2at^2
but i dunno if i am using the right equation....can someone help me plz?

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  1. If x is horizontal, no you are not using the right equation.
    Gravity has no horizontal component
    You horizontal speed, call it U is constant
    so
    X - Xo = U t
    In the vertical direction, fine with a = -g = -9.8 m/s

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  2. what is U...meaning acceleration?....

    so basically i would have

    1.80= -9.8t?...is that right?

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  3. Now to continue
    call the initial X0 = 0
    the final X = 1.42 m
    so
    1.42 = U t
    where U is that initial speed which remains the horizontal speed and t is time in the air
    NOW vertical direction
    call y upwards from the top of the table
    so Yo = 0
    Yfinal = - 1.90 m , the floor
    Vo = 0, no vertical speed originally, only the horizontal speed U
    so
    -1.90 = 0 - (1/2)(9.8) t^2
    t^2 = 1.9/4.9 = /0.388
    so
    t = 0.623 seconds in the air
    now back to get U
    1.42 = U t = U (.623)
    so
    U = 2.28 m/s

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  4. whoops, I used 1.90 instead of 1.80 for table height. You will have to redo the arithmetic :)

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  5. I called U the horizonal speed, and V the vertical speed.
    The point is that this is two problems.
    1. a constant speed horizontal problem where distance = horizontal rate * time
    2. a vertical problem where distance is initial vertical speed times time plus (1/2) a t^2

    The two problems are connected by the time in the air, t

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