A spherical cannon ball of diameter $28 \; cm$ is melted and cast into a right circular cone with base diameter $35 \; cm$. Find the height of the cone, correct to one decimal place.
Diameter of spherical cannon ball $= 28 \; cm$
$\therefore \;$ Radius of spherical cannon ball $= r = 14 \; cm$
Base diameter of the right circular cone $= 35 \; cm$
$\therefore \;$ Base radius of the right circular cone $= R = 17.5 \; cm$
Let the height of he cone be $= H$
$\because \;$ The spherical cannon ball is melted and recast into a right circular cone,
Volume of sphere $=$ Volume of right circular cone
i.e. $\;$ $\dfrac{4}{3} \pi r^3 = \dfrac{1}{3} \pi R^2 H$
$\implies$ $H = \dfrac{4 r^3}{R^2} = \dfrac{4 \times 14^3}{17.5^2} = 35.84$
$\therefore \;$ Height of cone $= 35.8 \; cm$ $\;$ (correct to one decimal place)