Respuesta :
Answer:
[tex]u_x=55.208\ m.s^{-1}[/tex]
Explanation:
Given:
horizontal distance form the point of shooting where the arrow hits ground, [tex]s=107\ ft[/tex] [tex]=32.614\ m[/tex]
angle below the horizontal form the point of release of arrow where it hits ground, [tex]\theta=3^{\circ}[/tex]
So the height above the ground from where the arrow was shot:
[tex]\tan3^{\circ}=\frac{h}{107}[/tex]
[tex]h=5.6076\ ft=1.71\ m[/tex]
- Since the arrow is shot horizontally so the initial vertical component of the velocity is zero ( [tex]u_y=0[/tex] ), we've the final vertical component of the velocity as:
[tex]v_y=\sqrt{2g.h}[/tex]
[tex]v_y=\sqrt{2\times 9.8\times 1.71}[/tex]
[tex]v_y=5.789\ m.s^{-1}[/tex]
Using equation of motion:
[tex]v_y=u_y+g.t[/tex]
where:
t = time taken
[tex]5.789=0+9.8\times t[/tex]
[tex]t=0.591\ s[/tex]
- Now the horizontal component of speed of the arrow (which remains constant throughout the motion by the Newton's first law of motion):
[tex]u_x=\frac{s}{t}[/tex]
[tex]u_x=\frac{32.614}{0.591}[/tex]
[tex]u_x=55.208\ m.s^{-1}[/tex]
