# Couple of kinematics problems

by Phantom8295
Tags: couple, kinematics
 P: 1,004 Let H be the height of the cliff. The function describing the rock's location is: $$y_{(t)} = \frac{1}{2}gt^2$$ From there you can find the time it takes the rock to reach the bottom of the cliff: $$y_{(t = t_f)} = \frac{1}{2}gt_f^2 = H$$ $$t_f = \sqrt{\frac{2H}{g}}$$ Now, how much time does it take sound to travel a distance of H (from the bottom of the cliff to your ears) at a speed of 340m/s? $$H = v_st_s$$ $$t_s = \frac{H}{v_s}$$ Finally, you know that the sum of these two time intervals should equal 3.4 seconds: $$t_f + t_s = 3.4$$ $$\sqrt{\frac{2H}{g}} + \frac{H}{v_s} = 3.4$$ If you let $$x = \sqrt{H}$$ you get this equation: $$\frac{x^2}{v_s} + x\sqrt{\frac{2}{g}} - 3.4 = 0$$ Now solve for x and then find H. I get H = 51.7 meters.