A rock is thrown off of the edge of a 9.8 meter high cliff with an initial horizontal velocity of 8.9 m/s.
It will land ____ m from the base of the vertical cliff.
Let dx = horizotal distance
Let dy = vertical distance
\(\begin{array}{|rcll|} \hline d_y &=& \frac12 \cdot g\cdot t^2 \qquad & | \qquad d_y = 9.8\ m \qquad g = 9.8\ \frac{m}{s^2}\\ 9.8\ m &=& \frac12 \cdot 9.8\ \frac{m}{s^2}\cdot t^2 \qquad & | \qquad : 9.8\\ 1\ m &=& \frac12 \cdot\frac{m}{s^2}\cdot t^2 \\ 1 &=& \frac12 \cdot\frac{1}{s^2}\cdot t^2 \qquad & | \qquad *2\\ 2s^2 &=& t^2 \\ \sqrt{2}\ s &=& t \\ \mathbf{t} & \mathbf{=} & \mathbf{\sqrt{2}\ s} \\ \hline \end{array} \)
The time the rock will land is \(t = \sqrt{2}\ s \)
\(\begin{array}{|rcll|} \hline d_x &=& v_0\cdot t \qquad & | \qquad v_0 = 8.9\ \frac{m}{s} \qquad t = \sqrt{2}\ s \\ d_x &=& 8.9\ \frac{m}{s}\cdot \sqrt{2}\ s \\ d_x &=& 8.9 \cdot \sqrt{2}\ m \\ d_x &=& 8.9 \cdot 1.41421356237\ m \\ d_x &=& 12.5865007051\dots \ m \\ \hline \end{array} \)
It will land 12.59 m from the base of the vertical cliff.
