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You have a friend who is an archer and she wants to know how fast the arrows leave her bow. You have her launch an arrow horizontally, 1.5 m above the ground, and find that the arrow lands 33 m away. What is the initial velocity of the arrow as it leaves the bow?

  • physics -

    horizontal: 33=V*time
    vertical: 0=1.5-1/2 9.8 t^2
    solve the first equation for time
    Put that in the second equation for t.
    Solve for V.

  • physics -

    Dvide 33 m by the time it takes the arrow to fall 1.5 meters. Call bhat time T.

    That time of flight T is obtainable from the relation:
    1.5 = (g/2)T^2.
    T = sqrt(3.0/g) = 0.553 s

    33 m/0.553s = ___ m/s

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