Answer:
The hang time is 2.04 seconds
Step-by-step 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
be the launching speed forming an angle
with the horizontal reference. The hang time (time the object remains in the air) is given by
![\displaystyle t_h=(2V_(oy))/(g)](https://img.qammunity.org/2020/formulas/physics/middle-school/kpgupapgyg03ikwk1ryq20anp5qe6vfe1i.png)
Since
![\displaystyle v_(oy)=v_o\ sin\theta](https://img.qammunity.org/2020/formulas/physics/middle-school/r5w8jdn6j2t14vg9t5o61550y84pdttb5b.png)
Then
![\displaystyle t_h=(2v_o\ sin\theta )/(g)](https://img.qammunity.org/2020/formulas/physics/middle-school/czwt68pmy5ibn0n7ur3c441h107dz2dxow.png)
We'll use the given values
![\displaystyle v_o=20\ m/s\ ,\ \theta =30^o](https://img.qammunity.org/2020/formulas/physics/middle-school/wf1ijy7xpiq55689a63n0yu3rj7shh2biq.png)
![\displaystyle t_h=(2(20)sin30^o)/(9,8)](https://img.qammunity.org/2020/formulas/physics/middle-school/1ngtlvg0e0wdqk87mnlxe304ovfmwhdz3y.png)
![\displaystyle t_h=2.04\ sec](https://img.qammunity.org/2020/formulas/physics/middle-school/ch9xvt1cq9wr6ufff9do66eu515lxc6fmx.png)
The hang time is 2.04 seconds