# Answer on Mechanics | Relativity Question for Christa

Question #20776

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?

A) what is the horizontal range of the arrow?

B) what is the maximum height of the arrow?

Expert's answer

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

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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