Michael Jordan’s vertical leap was measured to be 122 cm. What was his takeoff speed, and how long would he be in the air?
he take off speed was 5.1 and he was in the air for 1s
To calculate Michael Jordan's takeoff speed and the time he would be in the air, we can use the principles of physics. The relevant equation for vertical motion is:
vf^2 = vi^2 + 2as
Where:
- vf is the final velocity (zero when Michael reaches the highest point of his jump)
- vi is the initial velocity (takeoff speed)
- a is the acceleration due to gravity (-9.8 m/s^2 for objects near the Earth's surface)
- s is the displacement (vertical leap height, which is 122 cm or 1.22 meters)
Let's solve for vi (takeoff speed) first:
0 = vi^2 + 2(-9.8)(1.22)
vi^2 = 9.8(1.22)
vi^2 = 11.96
vi = √(11.96)
vi ≈ 3.46 m/s
Michael Jordan's takeoff speed would be approximately 3.46 meters per second.
Next, let's calculate the time he would be in the air. To do this, we'll use the equation:
vf = vi + at
Where:
- vf is the final velocity (zero when Michael reaches the highest point of his jump)
- vi is the initial velocity (takeoff speed)
- a is the acceleration due to gravity (-9.8 m/s^2 for objects near the Earth's surface)
- t is the time in seconds
0 = 3.46 + (-9.8)t
9.8t = 3.46
t = 3.46 / 9.8
t ≈ 0.35 seconds
Michael Jordan would be in the air for approximately 0.35 seconds.