Aregulation basketball has a 22 cm di-
ameter and may be approximated as a thin
spherical shell.
how long will it take a basketball starting
from rest to roll without slipping 1.1 m down
an incline that makes an angle of 49.0° with
the horizontal? the acceleration of gravity is
9.81 m/s2.
answer in units of s.

0.70 s

Explanation:

Potential energy = kinetic energy + rotational energy

mgh = ½ mv² + ½ Iω²

For a thin spherical shell, I = ⅔ mr².

mgh = ½ mv² + ½ (⅔ mr²) ω²

mgh = ½ mv² + ⅓ mr²ω²

For rolling without slipping, v = ωr.

mgh = ½ mv² + ⅓ mv²

mgh = ⅚ mv²

gh = ⅚ v²

v = √(1.2gh)

v = √(1.2 × 9.81 m/s² × 1.1 m sin 49.0°)

v = 3.13 m/s

The acceleration down the incline is constant, so given:

Δx = 1.1 m

v₀ = 0 m/s

v = 3.13 m/s

Find: t

Δx = ½ (v + v₀) t

t = 2Δx / (v + v₀)

t = 2 (1.1 m) / (3.13 m/s + 0 m/s)

t = 0.704 s

Rounding to two significant figures, it takes 0.70 seconds.

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.