Question

In 1989 during overtime in a high school basketball game in Erie, Pennsylvania, Chris Eddy threw a basketball with a velocity of 16.6 m/s at 50.0° and scored the winning basket.

(a) How long was the basketball in the air?

Answer 1

The basketball was in the air for approximately 2.587 seconds.

Step-by-step explanation:

In order to determine how long the basketball was in the air, we can use the horizontal motion and ignore air resistance. Since there is no horizontal acceleration, the time of flight in the air will only be determined by the vertical motion. We can use the vertical component of the initial velocity to find the time of flight. The initial vertical velocity can be calculated using the given velocity and angle:

Vertical velocity: vy = v * sin(θ)

Now we can use the vertical velocity to find the time of flight using the formula:

Time of flight: t = 2 * vy / g

where g is the acceleration due to gravity. Substituting the values gives:

t = 2 * (16.6 m/s * sin(50.0°)) / g

Using the known value for g (9.8 m/s2), we can calculate the time of flight:

t = 2 * (16.6 m/s * 0.766) / 9.8 m/s2 = 2 * 12.691 / 9.8 = 2.587 seconds

Therefore, the basketball was in the air for approximately 2.587 seconds.