When a slice of buttered toast is accidentally pushed over the edge of a counter, it rotates as it falls. If the distance to the floor is 80 cm and for rotation less than I rev, what are the (i) smallest and (ii) largest angular speeds that cause the toast to hit and then topple to be butter-side down?

The time it takes for the buttered toast to hit the floor is

$∆\mathrm{t} = \sqrt{\frac{2\mathrm{h}}{\mathrm{g}}} = \sqrt{\frac{2(8 \mathrm{m})}{10 \mathrm{m}/{\mathrm{s}}^2}} = 0.40\mathrm{s}$

The smallest angle turned for the toast to land butter side down is $∆{\mathrm{\theta }}_{\min } = \frac{1}{4}\mathrm{rev} =\frac{\mathrm{\pi }}{2}$. This corresponds to an angular speed of

${\mathrm{\omega }}_{\min } = \frac{∆{\mathrm{\theta }}_{\min }}{∆\mathrm{t}} = \frac{{\displaystyle \frac{\mathrm{\pi }}{2}}\mathrm{rad}}{0.4 \mathrm{s}} = \frac{5\mathrm{\pi }}{4}\mathrm{rad}/\mathrm{s}$

The largest angle (less than 1 revolution) turned for the roast to land butter-side down is $∆{\mathrm{\theta }}_{\max } = 0.75 \mathrm{rev} = \frac{3\mathrm{\pi }}{2}\mathrm{rad}.$ This corresponds to an angular speed of 

${\mathrm{\omega }}_{\max } = \frac{∆{\mathrm{\theta }}_{\max }}{∆\mathrm{t}} = \frac{{\displaystyle \frac{3\mathrm{\pi }}{2}}\mathrm{rad}}{0.4 \mathrm{s}} = \frac{15\mathrm{\pi }}{4}\mathrm{rad}/\mathrm{s}$