Answer: 2.17 s
Explanation:
The described situation is related to projectile motion (also known as parabolic motion). So, this kind of motion has a vertical component and a horizontal component; however, in this case we will only need the equation related to the horizontal displacement [tex]x[/tex]:
[tex]x=V_{o}cos \theta t[/tex]
Where:
[tex]x=45 m[/tex] is the arrow's horizontal displacement
[tex]V_{o}=800 m/s[/tex] is the arrow's initial velocity
[tex]\theta=75\°[/tex] is the angle
[tex]t[/tex] is the time the arrow is in the air
Isolating [tex]t[/tex]:
[tex]t=\frac{x}{V_{o}cos \theta}[/tex]
Solving with the given data:
[tex]t=\frac{45 m}{800 m/s cos(75\°)}[/tex]
[tex]t=2.17 s[/tex] This is the time the arrow is in the air
