Question

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. (a) what are the coordinates of the initial position of the rock? (enter your answers in m.) x_0 = 0 y_0 = 38.0 (b) what are the components of the initial velocity? (enter your answers in m/s.) v_0x = v_0y = what is the initial speed? m/s what is the initial direction of the rocks velocity? (c) write the equations for the x - and y - components of the velocity of the rock with time. (use the following as necessary: t. assume that v_x and v_y are in m/s and t is in seconds. do not include units in your answer.) if air resistance can be ignored, whats true about the x - component of the velocity? how does it relate to the initial v_0x you found in part (b)? m/s

Explanation:

Step1: Identify initial position

The origin is at the bottom of the cliff. The rock leaves the hand at a height of $y_0 = 38.0$ m above the ground - level at the bottom of the cliff and $x_0=0$ m. So the initial - position coordinates are $x_0 = 0$ m and $y_0=38.0$ m.

Step2: Analyze initial velocity components

The rock is thrown horizontally. In horizontal (x - direction), there is no acceleration ($a_x = 0$) and the initial velocity component in the x - direction $v_{0x}$ is non - zero. In the vertical (y - direction), the initial velocity component $v_{0y}=0$ m/s. Since it is thrown horizontally, if the initial speed is $v_0$, and the angle of projection $\theta = 0^{\circ}$ with respect to the x - axis. Using the equations $v_{0x}=v_0\cos\theta$ and $v_{0y}=v_0\sin\theta$, with $\theta = 0^{\circ}$, we have $v_{0x}=v_0$ and $v_{0y}=0$ m/s. Given the initial speed $v_0 = 16.5$ m/s, so $v_{0x}=16.5$ m/s and $v_{0y}=0$ m/s.

Step3: Find equations for velocity components

In the x - direction, since $a_x = 0$, the velocity - time equation $v_x=v_{0x}+a_xt$ gives $v_x = v_{0x}=16.5$ m/s (constant because $a_x = 0$). In the y - direction, the acceleration $a_y=-g=- 9.8$ m/s², and using the velocity - time equation $v_y=v_{0y}+a_yt$, we get $v_y=0 - gt=-9.8t$ m/s.

Answer:

(a) $x_0 = 0$ m, $y_0 = 38.0$ m
(b) $v_{0x}=16.5$ m/s, $v_{0y}=0$ m/s
(c) $v_x = 16.5$ m/s, $v_y=-9.8t$ m/s