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$V=πr^2\frac{h}{3}$

$A=πr^2$

$V=A\frac{h}{3}$

$42=42\frac{h}{3}$

$\boxed{h=3}$

Assuming that this is a right circular cone, the volume is given by $v=\dfrac{1}{3}A_b h$ where $A_b$ is the area of the base and $h$ is the height. Substituting,

$42=\dfrac{1}{3}(42)(h)$

$42$ cancels out

$\boxed{3=h}$