If a basketball player has a vertical leap of 1.29 m , then what is his takeoff speed and his hang time ( total time to move upwards to the peak and then return to the ground ) ?
time rising: 4.9t^2 = 1.29
double that for total hang time.
v = 9.8t
Hello sir thank you for answering but I can't seem to find how we got the v ( velocity ) to be as 9.8 ... Isn't that the acceleration
distance: s = 1/2 at^2
speed: v = at
To calculate the takeoff speed and hang time of a basketball player with a given vertical leap, we can use equations of motion. Here's how you can calculate them:
1. Takeoff Speed:
The takeoff speed can be determined using the equation for vertical motion:
v = sqrt(2 * g * h)
where:
v is the takeoff speed,
g is the acceleration due to gravity (approximately 9.8 m/s^2), and
h is the vertical leap (1.29 m).
Substituting the values into the equation:
v = sqrt(2 * 9.8 * 1.29)
v ≈ sqrt(25.372)
Therefore, the takeoff speed is approximately 5.04 m/s.
2. Hang Time:
To calculate the total time for the player to move upwards to the peak and return to the ground (hang time), you can use the equation for the time of flight for vertical motion:
t = sqrt((2 * h) / g)
where:
t is the total time of flight,
h is the vertical leap (1.29 m), and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Substituting the values into the equation:
t = sqrt((2 * 1.29) / 9.8)
t ≈ sqrt(0.2633)
Therefore, the hang time is approximately 0.513 seconds.
So, the takeoff speed of the basketball player is approximately 5.04 m/s, and the hang time (total time to move upwards and return to the ground) is approximately 0.513 seconds.