\[\boxed{\text{12\ (12).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[h = S \cdot \frac{1}{2} = vt \cdot \frac{1}{2};\ \ \ v = 0,75:\]
\[h(t) = 0,75t \cdot sin30{^\circ} =\]
\[= \frac{3}{4}\ t \cdot \frac{1}{2} = \frac{3}{8}\text{t.}\]
\[\textbf{а)}\ t = 2,25\ мин:\ \]
\[h(2,25) = \frac{3}{8} \cdot 2,25\ (мин) =\]
\[= \frac{3}{8} \cdot 135\ (сек) = 50,625\ (м).\]
\[Ответ:50,625\ м.\]
\[\textbf{б)}\ h = 60\ м:\ \]
\[60 = \frac{3}{8}\text{t\ }\]
\[t = \frac{60 \cdot 8}{3} = 160\ с.\]
\[Ответ:160\ с.\]
\[\boxed{\text{12.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ 200;\]
\[\textbf{б)} - 2;\ - 100;\]
\[\textbf{в)}\ 0;200;\]
\[\textbf{г)} - 2;0;8,83;\ \frac{1}{48};\ 200;\ - 100;\]
\[\ \frac{2}{3};\ - 5,12;\ - \frac{3}{7};0,0002;\]
\[\textbf{д)}\ \sqrt{2};\ \pi;\ - \sqrt{11};\ \]
\[\textbf{е)}\ \sqrt{2};\ \pi;\ - \sqrt{11};\ - 2;0;8,83;\]
\[\ \frac{1}{48};\ 200;\ - 100;\ \frac{2}{3};\ - 5,12;\ \]
\[- \frac{3}{7};0,0002.\]