a boulder moving horizontally at 15.7 m/s on flat ground rolls off of a 25.3 m high cliff. how far from the…

Explanation:

Step1: Calculate time of fall

The vertical - motion of the boulder is a free - fall motion. The vertical displacement $y = h=25.3$ m, the initial vertical velocity $v_{0y}=0$ m/s, and the acceleration due to gravity $g = 9.8$ m/s². Using the equation $y=v_{0y}t+\frac{1}{2}gt^{2}$, since $v_{0y} = 0$, we have $y=\frac{1}{2}gt^{2}$. Solving for $t$ gives $t=\sqrt{\frac{2y}{g}}$. $t=\sqrt{\frac{2\times25.3}{9.8}}\text{ s}$

Step2: Calculate horizontal displacement

The horizontal motion of the boulder is a uniform - motion with constant velocity because there is no horizontal acceleration ($a_x = 0$). The horizontal velocity $v_x=15.7$ m/s. The horizontal displacement $\Delta x$ is given by the equation $\Delta x=v_x t$. Substitute $t=\sqrt{\frac{2\times25.3}{9.8}}$ into $\Delta x = v_x t$. $\Delta x=15.7\times\sqrt{\frac{2\times25.3}{9.8}}$ $\Delta x=15.7\times\sqrt{\frac{50.6}{9.8}}\approx15.7\times\sqrt{5.163}\approx15.7\times2.272\approx35.7$ m

Answer:

$35.7$