Question

(II) The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about 120 cm. (a) What is their initial “launch” speed off the ground? (b) How long are they in the air?

Answer 1

Answer:

The answer is (a) 4.85 m/s (b) 0.495 s

Explanation:

Let u = the vertical launch speed.

Neglect air resistance, and g = 9.8 m/s².

At the maximum height of h = 120 cm = 1.2 m, the vertical velocity is zero.

Therefore

(u m/s)² - 2*(9.8 m/s²)*(1.2 m) = 0

u² = 23.52 (m/s)²

u = 4.85 m/s

The time to attain maximum height is one half of the travel time.

If t = time to attain maximum height, then

u - gt = 0

4.85 - 9.8 t = 0

t = 4.85/9.8 = 0.495 s

Therefore the travel time is 2 x 0.0495 = 0.99 s