Question

Two stones are thrown simultaneously, one straight upward from the base of a cliff and the other straight downward from the top of the cliff. The height of the cliff is 6.55 m. The stones are thrown with the same speed of 9.42 m/s. Find the location (above the base of the cliff) of the point where the stones cross paths.

D =

Answer 1

Question #14618

Let the coordinates of the stone, thrown downward be $y_{1}$ and the stone, thrown upward be $y_{2}$ . The general equation for a motion of an accelerated object is $y = y_{0} + v_{y}t + \frac{g t^{2}}{2}$ . Hence, the equations for $y_{1}, y_{2}$ are: $y_{1} = h - v_{0}t - \frac{g t^{2}}{2}, y_{2} = v_{0}t - \frac{g t^{2}}{2}$ , where $h$ is the height of cliff. In order to find the position, where the stones cross, $y_{1} = y_{2} \Rightarrow 2v_{0}t = h, t = \frac{h}{2v_{0}}$ - this is the interval time from the beginning of motion, when stones cross. Finally, the height from the base of cliff is $D = \frac{v_{0}h}{2v_{0}} - \frac{g h^{2}}{8v_{0}^{2}} = 2.67m$ (substituting the time we found into $y_{2}$ ).