Question

13. Bari kicked a ball and its leaves the ground at an angle 30° with a velocity of 20 m/s.

Answer 1

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 $v_o$ be the launching speed forming an angle $\theta$ with the horizontal reference. The hang time (time the object remains in the air) is given by

$\displaystyle t_h=\frac{2V_{oy}}{g}$

Since

$\displaystyle v_{oy}=v_o\ sin\theta$

Then

$\displaystyle t_h=\frac{2v_o\ sin\theta }{g}$

We'll use the given values

$\displaystyle v_o=20\ m/s\ ,\ \theta =30^o$

$\displaystyle t_h=\frac{2(20)sin30^o}{9,8}$

$\displaystyle t_h=2.04\ sec$

The hang time is 2.04 seconds