A soccer ball is kicked with a speed of 9.80 m/s at an angle of 25.0 degrees above the horizontal. If the ball lands at the same level from which it was kicked, how long was it in the air?

figure the vertical velocity, then knowing the final and initial heights, solve for time in air.

Twice the time that it takes the vertical velocity component (9.80 sin 25, m/s) to become zero. The same time is required to go up as come back down. Hence the factor of two.

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To find the time the soccer ball was in the air, you can use the kinematic equations of motion. In this case, we can use the equation of motion that relates the horizontal distance traveled, the initial velocity in the horizontal direction, and the time of flight.

The horizontal distance traveled by the soccer ball can be found using the equation:

\(d = v * t\)

where d is the horizontal distance, v is the initial horizontal velocity, and t is the time of flight.

In this case, the initial horizontal velocity can be found using the trigonometric function cosine, since the angle of 25.0 degrees is above the horizontal. The equation for the initial horizontal velocity (v) is:

\(v = v_0 * \cos(\theta)\),

where v_0 is the initial velocity of the soccer ball and theta is the angle above the horizontal.

The initial velocity in the horizontal direction (v_0) is equal to the initial speed of the soccer ball, which is given as 9.80 m/s.

So, substituting the values, we have:

\(v = 9.8 * \cos(25.0^\circ)\).

Now, we can find the horizontal distance (d) by multiplying the initial horizontal velocity (v) with the time of flight (t). Since the ball lands at the same level from which it was kicked, the vertical displacement is zero.

Finally, we can find the time of flight by rearranging the equation and solving for t:

\(t = \frac{d}{v}\).

Substituting the values, we can now calculate the time of flight.