a basketball player achieves a hang time of 0.996 s in dunking the ball. What vertical height will he attain?
Hint: He spends 0.498 s coming back down.
H = (g/2) t^2
Use t = 0.498 s
H = 1.21 meters = 3.97 feet
That height applies to his center of mass, relative to where it is when he lands. His outstretched hand, containing the ball, will be much higher above the floor at that time.
Assuming the player's CM is 3 feet above the floor and the ball in the hand is 4 feet above the center of mass, the ball is about 11 feet above the floor at max height, which is sufficient for a dunk.
To calculate the vertical height attained by the basketball player, we can use the equation of motion for vertical motion:
h = (gt^2) / 2
Where:
h = vertical height attained
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time of hang time (0.996 s)
Plugging in the given values:
h = (9.8 * (0.996)^2) / 2
Calculating the equation:
h = (9.8 * 0.992016) / 2
h ≈ 4.87465 meters
Therefore, the basketball player will attain a vertical height of approximately 4.87 meters.
To determine the vertical height the basketball player will attain, we can use the equation of motion:
h = (1/2) * g * t^2
Where:
h is the vertical height
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time (hang time) in seconds
Substituting the given values, we have:
h = (1/2) * 9.8 m/s^2 * (0.996 s)^2
Calculating this equation will give us the answer:
h = (1/2) * 9.8 m/s^2 * 0.992016 s^2
h = 4.812 m
Therefore, the basketball player will attain a vertical height of approximately 4.812 meters.