Speed of arrow when it leaves the bow?

Question

A 0.05kg arrow is launched vertically into the air. The bowstring exerts an average force of 53N on the arrow over a distance of 0.75m.
a) with what speed does the arrow leave the bow
b) how high will the arrow travel. neglect air resistance

thanks!

Answer 1

a)
v^2-0 = 2as
v = sqrt(2*0.75*(53 - 0.05*9.8)/0.05)

b)
s = (v^2)/2a)
s = (2*0.75*(53 - 0.05*9.8)/0.05)/(2*9.8)

Answer 2

a=9.8 m/s^2
F=53 kg*m/s^2 =53N
d=0.75 m

Energy=Force*dist=mass*vel^2
F*d=m*v^2
v^2=F*d/m
v=sqrt(F*d/m)= sqrt(53*0.75/0.05)~28.196 meters/second at launch

v=v0+at
arrow will have stopped when v=0 so:
0=v0+a*t
t=-v0/a=-(28.196/-9.8)=~2.877 seconds after launch

s=(1/2)a*t^2+v0*t+s0
substitute v0=28.196,a=-9.8,s0=0 and t=2.877 and you will get:
s=40.562 meters

Answer 3

335mph