Question

Michael jordan was said to have a hang-time of 3.0 seconds. use kinematic equations to determine the height to which he could leap

Answer 1

Using kinematics, it's calculated that Michael Jordan could potentially leap to a height of approximately 11.025 meters with a hang-time of 3.0 seconds.

Step-by-step explanation:

To determine the height to which Michael Jordan could leap with a hang-time of 3.0 seconds, we will use the kinematic equations for projectile motion. The total hang-time is the time it takes to reach the peak of the jump and the time to come back down, which are equal. Therefore, the time to reach the peak (t) is half of the total hang-time.

t = hang-time / 2 = 3.0s / 2 = 1.5s

Using the kinematic equation for displacement (h) where initial velocity is 0 at the peak, acceleration due to gravity (g) is -9.8 m/s², and time (t) is 1.5s, we get:

h = v * t + (1/2) * g * t²

h = 0 * 1.5s + (1/2) * (-9.8 m/s²) * (1.5s)²

h = -0.5 * 9.8 m/s² * 2.25s²

h = -0.5 * 9.8 m/s² * 2.25 s² = -11.025 m

Because height cannot be negative in this context, we take the absolute value:

h = 11.025 m

Thus, using kinematic equations, Michael Jordan could potentially leap to a height of approximately 11.025 meters.

Answer 2

The hang time is 3 seconds, which is the time of travel off the ground.
The maximum height is attained in 3/2 = 1.5 s.

Acceleration due to gravity = 9.8 m/s
Wind resistance is ignored.

At maximum height, the vertical velocity is zero.
If u is the vertical launch velocity, then
(u m/s) - (9.8 m/s²)*(1.5 s) = 0
u = 14.7 m/s

The maximum height is
h = (14.7 m/s)*(1.5 s) - 0.5*(9.8 m/s²)*(1.5 s)²
=11.025 m

Answer: 11.0 m (nearest tenth)