Answer:
The hang time is 2.04 seconds
Explanation:
2-D Motion
It referred to as a situation where an object is launched in such a way it describes a curve, reaches a top height and then returns to ground level after traveling a certain distance x away from the launch point.
Let [tex]v_o[/tex] be the launching speed forming an angle [tex]\theta[/tex] with the horizontal reference. The hang time (time the object remains in the air) is given by
[tex]\displaystyle t_h=\frac{2V_{oy}}{g}[/tex]
Since
[tex]\displaystyle v_{oy}=v_o\ sin\theta[/tex]
Then
[tex]\displaystyle t_h=\frac{2v_o\ sin\theta }{g}[/tex]
We'll use the given values
[tex]\displaystyle v_o=20\ m/s\ ,\ \theta =30^o[/tex]
[tex]\displaystyle t_h=\frac{2(20)sin30^o}{9,8}[/tex]
[tex]\displaystyle t_h=2.04\ sec[/tex]
The hang time is 2.04 seconds
