The hang-time of a basketball player who jumps a vertical distance of 2 feet (0.61 m) is about 0.70 second. What will be the hang-time if the player reaches the same height while jumping 3 feet (0.91 m) horizontally?
Cubv
To determine the hang-time of the player, we can use the standard physics equation for vertical motion:
hang-time = 2 * time of flight
Where "time of flight" refers to the time it takes for the player to reach the peak of their jump. We can find the time of flight using the equation:
time of flight = (2 * vertical distance) / acceleration due to gravity
The acceleration due to gravity is a constant value of approximately 9.8 m/s^2. Given that the player jumps a vertical distance of 2 feet (0.61 m), we can calculate the time of flight:
time of flight = (2 * 0.61) / 9.8
= 0.122 seconds
Now, let's find the hang-time for this vertical jump:
hang-time = 2 * time of flight
= 2 * 0.122
= 0.244 seconds
Now, let's determine the hang-time when the player reaches the same height while jumping 3 feet (0.91 m) horizontally. Since the horizontal motion does not affect the vertical motion, the time of flight will remain the same. Therefore, the hang-time will also be:
hang-time = 2 * time of flight
= 2 * 0.122
= 0.244 seconds
Thus, the hang-time will be approximately 0.244 seconds when the player reaches the same height while jumping 3 feet horizontally.