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?

Question

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

Answer 1

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