a student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of vO= 12.5 m/s. the cliff is h=38.0 m above a flat, horizontal beach as shown in the figure. (c) write the equations for the c- and y-components of the velocity of the stone with time. (use the following as necessary: t. let the variable t be measured in seconds. do not include units in your answer.) (d) write the equations for the position of the stone with time, using the coordinates in the figure. (use the following as necessary:t. let the variable t be measured in seconds. do not state units in your answer.)

To solve this problem, we need to consider the motion of the stone in two dimensions: horizontally (c-component) and vertically (y-component).

(c) The equation for the c-component of velocity remains constant since the stone is thrown horizontally. Therefore, the equation for the c-component of velocity of the stone with time is:

v_c(t) = v_O

where v_O is the initial velocity in the horizontal direction (12.5 m/s).

The equation for the y-component of velocity can be determined using the kinematic equation for horizontal motion:

v_y(t) = v_{y0} - g \cdot t

where v_{y0} is the initial vertical velocity (0 m/s) and g is the acceleration due to gravity (-9.8 m/s^2).

(d) The equations for the position of the stone with time can be obtained by integrating the velocity equations with respect to time.

The equation for the horizontal position (c-coordinate) of the stone with time is:

x(t) = v_O \cdot t + x_0

where x_0 is the initial horizontal position (0 m).

The equation for the vertical position (y-coordinate) of the stone with time is:

y(t) = v_{y0} \cdot t - \frac{1}{2} \cdot g \cdot t^2 + y_0

where y_0 is the initial vertical position (38.0 m).

Answer:
(c) v_c(t) = v_O
v_y(t) = v_{y0} - g \cdot t

(d) x(t) = v_O \cdot t + x_0
y(t) = v_{y0} \cdot t - \frac{1}{2} \cdot g \cdot t^2 + y_0