9000022901

An arrow has been shot at the angle ${60}^{\circ }$ at the speed $10\mathrm{m}{\mathrm{s}}^{-1}$. Find the time when the height equals to the horizontal distance from the take-off point. Hint: The position is given by the equations $x=v_{0}t\cdot \cos \alpha$, $y=v_{0}t\cdot \sin \alpha -\frac{1}{2}g{t}^2$. Use $g=10\mathrm{m}{\mathrm{s}}^{-2}$ as an acceleration of gravity.