A basketball star covers 2.60 m horizontally in a jump to dunk the ball. His motion through space can be modeled precisely as that of a particle at his center of mass. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.90 m above the floor and is at elevation 0.920 m when he touches down again.(a) Determine his time of flight (his "hang time").

To figure out the basketball player's hang time, we need to understand a bit about how gravity works. When the player jumps into the air, gravity pulls him back down. This happens at a constant rate, which on Earth is about 9.8 meters per second squared.

First, we need to find out how long the player was rising and falling. We know that the player's center of mass was 1.02 meters above the ground when he jumped, and it reached a maximum height of 1.90 meters. That's a difference of 1.90 - 1.02 = 0.88 meters.

Gravity pulls him down at 9.8 meters per second squared, but he's going up first, so we'll call it -9.8 meters per second squared. To find out how long he was rising, we can use the equation of motion: final velocity = initial velocity + (acceleration * time). When he reaches the highest point of his jump, his final velocity is 0. So, we can rearrange the equation to solve for time: time = (final velocity - initial velocity) / acceleration.

We don't know the player's initial velocity, but we can find it using another equation of motion: final velocity^2 = initial velocity^2 + 2*(acceleration * distance). Plugging in the numbers we know, we get 0 = initial velocity^2 + 2*(-9.8 * 0.88). Solving for initial velocity, we get about 4.16 meters per second.

Now we can find the time it took for the player to rise: time = (0 - 4.16) / -9.8 = about 0.42 seconds.

The player also falls after reaching the maximum height. The time it takes to fall is the same as the time it took to rise, so the total hang time is 0.42 seconds * 2 = 0.84 seconds. So, the basketball player's hang time is approximately 0.84 seconds.