Question

rock a is thrown horizontally off of a cliff with a velocity of 20 m/s. the rock lands 60 m from the base of the cliff. according to the information and diagram above, what is the height of the cliff? 60.0 m 44.1 m 10.1 m 210 m

Explanation:

Step1: Find the time of flight

The horizontal - motion is a uniform - motion with $v_x = 20\ m/s$ and $x = 60\ m$. Using the formula $x=v_x t$, we can solve for $t$.
$t=\frac{x}{v_x}=\frac{60}{20}=3\ s$

Step2: Find the height of the cliff

The vertical - motion is a free - fall motion with initial vertical velocity $v_{0y}=0\ m/s$, acceleration $a = g=9.8\ m/s^{2}$, and time $t = 3\ s$. Using the formula $y=v_{0y}t+\frac{1}{2}gt^{2}$, since $v_{0y}=0\ m/s$, we have $y=\frac{1}{2}gt^{2}$.
$y=\frac{1}{2}\times9.8\times3^{2}=\frac{1}{2}\times9.8\times9 = 44.1\ m$

Answer:

$44.1\ m$