Question

A rebounder in v basketball has a vertical leap of 1.2 meter. What is his launch speed and "hang time"?

Answer 1

Answer on Question #38205 – Physics - Mechanics | Kinematics | Dynamics

A rebounder in v basketball has a vertical leap of 1.2 meter. What is his launch speed and "hang time"?

Solution:

S = 1.2m – height of vertical leap

V = 0 – velocity at the top (turning point)

Acceleration due to gravity, g = -9.81 $\frac{\text{m}}{\text{s}^2}$ (since motion is in opposite direction as gravity)

Law of energy conservation:

$$\frac{\mathrm{m V}^2}{2} = \frac{\mathrm{m U}^2}{2} + \mathrm{m g S}$$ $$V^2 = U^2 + 2 \mathrm{g S}$$ $$U = \sqrt{-2 \mathrm{g S}} = \sqrt{2 \cdot 9.81 \frac{\mathrm{m}}{\mathrm{s}^2} \cdot 1.2 \mathrm{m}} = 4.9 \frac{\mathrm{m}}{\mathrm{s}}$$

Hang time (assuming hang time is the total time from launch to land):

After land, displacement from launch to land is equal to zero:

$$0 = U t_h + \frac{g t_h^2}{2}$$ $$0 = 4.9 t_h - 4.9 t_h^2$$ $$t_h = 1 \mathrm{s}$$

Answer: launch speed is equal to 4.9 $\frac{\text{m}}{\text{s}}$; "hang time" is equal to 1s.