physics

what is the initial velocity of this equation.

.92m= 3.10 times .43 + 1/2 times 9.8 m/s times .43 squared

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asked by lisa
  1. This looks like the general equation equation of motion in a gravitational field,
    y = Yo + Vo t + (1/2) g t^2

    The initial velocity is Vo = 3.1 m/s.

    It appears to be an equation for vertical position (measured positive downward) at time t = 0.43 s. 0.92 m is the position at that time.

    The units of the "9.8" should be m/s^2, not m/s.

    You seem to have left out Yo, the term for the position at t=0, because the two sides of your equation are not equal. Do the numbers.

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    posted by drwls

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