Posted by **katelyn** on Friday, October 8, 2010 at 7:43pm.

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 -
**bobpursley**, Friday, October 8, 2010 at 7:51pm
horizontal: 33=V*time

vertical: 0=1.5-1/2 9.8 t^2

solve the first equation for time

time=33/V

Put that in the second equation for t.

Solve for V.

- physics -
**drwls**, Friday, October 8, 2010 at 7:51pm
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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