Question

1. 10 / 11 points a person stands at the edge of a cliff and throws a rock horizontally over the edge with a speed of v0 = 16.5 m/s. the rock leaves his hand at a height of h = 38.0 m above level ground at the bottom of the cliff, as shown in the figure. note the coordinate system in the figure, where the origin is at the bottom of the cliff, directly below where the rock leaves the hand. details my notes previous answers ask your teacher practice another

Explanation:

Step1: Analyze vertical - motion

The rock is in free - fall in the vertical direction. The initial vertical velocity $v_{0y}=0\ m/s$, the acceleration due to gravity $g = 9.8\ m/s^{2}$, and the vertical displacement $y=- 38.0\ m$ (taking downwards as negative). We use the equation $y = v_{0y}t+\frac{1}{2}at^{2}$. Since $v_{0y} = 0\ m/s$, the equation simplifies to $y=\frac{1}{2}at^{2}$, where $a=-g$. So, $-38.0=-\frac{1}{2}\times9.8t^{2}$.

Step2: Solve for time $t$

Rearranging the equation $-38.0 =-\frac{1}{2}\times9.8t^{2}$ for $t$, we get $t=\sqrt{\frac{2\times38.0}{9.8}}$.
\[t=\sqrt{\frac{76}{9.8}}\approx\sqrt{7.755}\approx2.78\ s\]

Step3: Analyze horizontal - motion

In the horizontal direction, there is no acceleration ($a_x = 0$), and the initial horizontal velocity $v_{0x}=16.5\ m/s$. The horizontal displacement is given by $x = v_{0x}t$.

Step4: Calculate horizontal displacement

Substitute $v_{0x}=16.5\ m/s$ and $t = 2.78\ s$ into the equation $x = v_{0x}t$. So, $x=16.5\times2.78 = 45.87\ m$.

Answer:

$45.9\ m$ (rounded to one decimal place)