Question

An arrow is fired horizontally from the top of a 40.0-m vertical cliff and lands 125 m away. Therefore the arrow was fired at _____ m/s.

Answer 1

Question 30758

The law of motion is as follows: $\nu_{x} = \nu_{0x} = \nu_{0}$ (according to conditions of the task, object was dropped only with horizontal initial velocity, let us denote it $\nu_{0}$ ). For vertical part of velocity: $\nu_{y} = \nu_{0y} - gt = -gt$ (because $\nu_{0y} = 0$ ).

Hence, horizontal distance, traveled by object is $S_{x} = v_{0} \cdot t$ , from where $v_{0} = \frac{S_{x}}{t}$ ( $S_{x} = 125m$ ).

Time might be found from vertical law of motion $S_{y} = \frac{g t^{2}}{2} \Rightarrow t = \sqrt{\frac{2 S_{y}}{g}}$ ( $S_{y} = 40m$ ).

Plugging the latter formula into formula for $\nu_{0}$ , obtain $\nu_{0} = S_{x}\frac{\sqrt{g}}{\sqrt{2S_{y}}} = 43.8\frac{m}{s}$ .