Question

18 a rock is kicked horizontally at a speed of 8 m/s from the edge of a cliff. the rock strikes the ground 65 m from the foot of the cliff of height h as sugge resistance. what is the height of the cliff? a 166 m b 323 m c 651 m d 4 m e 519 m

Explanation:

Step1: Find time of flight

The horizontal - motion is a uniform - motion with $v_x = 8\ m/s$ and $x = 65\ m$. Using the formula $x=v_x t$, we can find the time of flight $t$.
$t=\frac{x}{v_x}=\frac{65}{8}=8.125\ s$

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 $t = 8.125\ s$. Using the formula $y=v_{0y}t+\frac{1}{2}gt^{2}$, and since $v_{0y} = 0\ m/s$, we have $y=\frac{1}{2}gt^{2}$.
$y=\frac{1}{2}\times9.8\times(8.125)^{2}=\frac{1}{2}\times9.8\times66.015625\approx323\ m$

Answer:

B. 323 m