A Floating Basketball

Classical Mechanics Level 2

A basketball 24 cm diameter and 640 g weight, is floating in a pool filled with fresh water.

What is the distance (cm) from the top of the ball to the water level?

Assumptions:

  • The density of fresh water is 1 g / c m 3 . \SI[per-mode=symbol]{1}{\gram\per\centi\meter\cubed}.


The answer is 19.6.

Ahmed Almubarak by Ahmed Almubarak , 38, Iraq

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1 solution

Ahmed Almubarak
Jun 3, 2020

As a result of Archimedes principle "Weight of the ball = Weight of displaced fluid"

Weight of the displaced water = 640 g

Volume of the displaced water = 640 g 1 g / c m 3 = 640 c m 3 =\frac { 640 g}{ 1 g /{ cm }^{ 3 } } =640 { cm }^{ 3 }

Volume of the ball section immersed in water ( Vb ) = Volume of the displaced water = 640 c m 3 640{ cm }^{ 3 }

V b = π 3 h 2 ( 1.5 D − h ) { V }_{ b }=\frac { \pi }{ 3 } { h }^{ 2 }(1.5D-h)

640 = π 3 h 2 ( 1.5 ∗ 24 − h ) 640=\frac { \pi }{ 3 } { h }^{ 2 }(1.5*24-h) ------- Solve for h then h= 4.397599 cm

H = 24 − 4.397599 ≃ 19.6 c m H=24-4.397599\simeq \boxed { 19.6 cm }

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