Answer to Question #20776 in Mechanics | Relativity for Christa
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?
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
A. horizontal range is
v(hor.)*t = 50*cos(30)*5.1 = 220 m
B.Height of arrow is
h=v(vert)^2 / (2g) = 31.8 m