\(1\ yd=3\ ft\to1\ ft=\dfrac{1}{3}\ yd\\\\1\ h=60\ min\\\\therefore\\\\90\ \dfrac{ft}{h}=90\cdot\dfrac{\frac{1}{3}\ yd}{60\ min}=\dfrac{90\cdot\frac{1}{3}}{60}\ \dfrac{yd}{min}=\dfrac{30}{60}\ \dfrac{yd}{min}=\dfrac{1}{2}\ \dfrac{yd}{min}=\boxed{0.5\ \frac{yd}{min}}\)