Answer:
The correct option is 1. 720 m
Explanation:
Projectile Motion
When an object is launched in free air (no friction) with an initial speed vo at an angle [tex]\theta[/tex], it describes a curve which has two components: one in the horizontal direction and the other in the vertical direction. The data provided gives us the initial conditions of the survival package's launch.
[tex]\displaystyle V_o=250\ m/s[/tex]
[tex]\displaystyle \theta =-37^o[/tex]
The initial velocity has these components in the x and y coordinates respectively:
[tex]\displaystyle V_{ox}=250\ cos(-37^o)=199.7\ m/s[/tex]
[tex]\displaystyle V_{oy}=250\ sin(-37^o)=-150.5\ m/s[/tex]
And we know the plane has an altitude of 600 m, so the package will reach ground level when:
[tex]\displaystyle y=-600\ m[/tex]
The vertical distance traveled is given by:
[tex]\displaystyle y=V_{oy}\ t-\frac{g\ t^2}{2}=-600[/tex]
We'll set up an equation to find the time when the package lands
[tex]\displaystyle -150.5t-4.9\ t^2=-600[/tex]
[tex]\displaystyle -4.9\ t^2-150.5\ t+600=0[/tex]
Solving for t, we find only one positive solution:
[tex]\displaystyle t=3.6\ sec[/tex]
The horizontal distance is:
[tex]\displaystyle x=V_{ox}.t=199.7\times3.6=720\ m[/tex]
The correct option is 1. 720 m
