Posted by **Lindsay** on Wednesday, January 16, 2008 at 10:38pm.

An archer shoots an arrow that leaves the bow at an angle of 42 degrees above the horizontal. The arrow hits a target located 135 m away and 23 m above the height from which the arrow was shot. Given that the mass of the arrow is .125 kg and that it remained in contact with the string 1.28 m, what was the force applied to the arrow by the string of the bow?

Can someone at least tell me where to begin? I'm lost when it comes to these types of problems.

- Physics -
**Damon**, Thursday, January 17, 2008 at 4:30am
vertical direction equations:

v = Vo sin 42 -9.8 t

or

v = .669 Vo - 9.8 t

h = Ho + .669 Vo t - 4.9 t^2

Horizontal direction equation

x = (Vo cos 42 ) t

or

x = .743 Vo t

Now horizontal fact

135 = .743 Vo t

And vertical

23 = 0 + ,669 Vo t - 4.9 t^2

from horizontal equation

Vo t = 182

use that in vertical equation

23 = .669 (182) - 4.9 t^2

4.9 t^2 = 98.8

t = 4.49 seconds in air

so Vo = 182/4.49

Vo = 40.5 m/s

Now impuse = change of momentum

F d = m Vo

F (1.28) = .125 (40.5)

solve for F

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