physics
posted by katelyn .
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?

horizontal: 33=V*time
vertical: 0=1.51/2 9.8 t^2
solve the first equation for time
time=33/V
Put that in the second equation for t.
Solve for V. 
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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