Asked by J_nin
                An archer needs to hit a bull’s eye on a target at eye level 60.0 ft away. If the archer fires the arrow from eye level with a speed of 47.0 ft/s at an angle of 25 degree above the horizontal, will the arrow hit the target?
            
            
        Answers
                    Answered by
            J_nin
            
    convert 47 ft/s to m/s = 14.3m/s
V = 14.3 m/s
Angle = 25 degree
|Vx|=|V|cos(25)
|Vx|=|14.3 cos(25)= 13 m/s
     
|Vy|=|V|sin(25)
|Vy|=|14.3 sin(25)= 6.04 m/s
t= Vy x 2 /9.8
t= 6.04m/s x 2 /9.8 m/s^2 = 1.23 s
SX = VX * t
then convert it back to feet
so the arrow will miss the target as the arrow travel about 52.3 feet
   
    
V = 14.3 m/s
Angle = 25 degree
|Vx|=|V|cos(25)
|Vx|=|14.3 cos(25)= 13 m/s
|Vy|=|V|sin(25)
|Vy|=|14.3 sin(25)= 6.04 m/s
t= Vy x 2 /9.8
t= 6.04m/s x 2 /9.8 m/s^2 = 1.23 s
SX = VX * t
then convert it back to feet
so the arrow will miss the target as the arrow travel about 52.3 feet
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