Solving Physics Problem on Ball Motion

Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the same instant, the other student throws a ball, b2, vertically upward at the same speed. The second ball just misses the balcony on the way down.

(a) What is the difference in the time the balls spend in the air?

The time difference is 2.83 seconds.

(b) What is the velocity of each ball as it strikes the ground?

Velocity of the ball thrown downward is 23.8 m/s.

Velocity of the ball thrown upward is 23.8 m/s.

(c) How far apart are the balls 0.800 s after they are thrown?

The distance between the balls after 0.8 seconds is 22.24 meters.

Explanation:

Part (a): The time difference is calculated by considering the motion of the ball thrown upward and downward. The time difference is found to be 2.83 seconds.

Part (b): The final velocity of both balls is calculated to be 23.8 m/s based on their initial speed and displacement.

Part (c): The distance between the two balls at 0.8 seconds is determined using the relative speed of the balls, resulting in a distance of 22.24 meters.

Two students are on a balcony 19.1 m above the street. One student throws a ball, b1, vertically downward at 13.9 m/s. At the same instant, the other student throws a ball, b2, vertically upward at the same speed. The second ball just misses the balcony on the way down.

1. What is the difference in the time the balls spend in the air?
2. What is the velocity of each ball as it strikes the ground?
3. How far apart are the balls 0.800 s after they are thrown?

1. The time difference is 2.83 seconds.
2. Velocity of the ball thrown downward is 23.8 m/s. Velocity of the ball thrown upward is 23.8 m/s.
3. The distance between the balls after 0.8 seconds is 22.24 meters.