an arrow is shot from a bow at a 30 degree angle and a velocity of 50 m/s.
A) what is the horizontal range of the arrow?
B) what is the maximum height of the arrow?
1
Expert's answer
2012-12-20T09:17:26-0500
The vertical initial speed is 50*sin(30) = 50/2 = 25 Hence, if flight will last time t, the equation for vertivalvelocity is 2v(vert.)=gt, t =2v/g = 50/9.8 = 5.1 sec
so, A. horizontal range is v(hor.)*t = 50*cos(30)*5.1 = 220 m B.Height of arrow is mgh=m v(vert)^2/2 h=v(vert)^2 / (2g) = 31.8 m
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