Question

# 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. $use the following as necessary:t. let the variable t be measured in seconds. do not state units in your answer.$

118

likes
591 views

## Answer to a math question 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. $use the following as necessary:t. let the variable t be measured in seconds. do not state units in your answer.$

Hank
4.8
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 $38.0 m$.

$c$ 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
Frequently asked questions $FAQs$
What is the result of applying the chain rule to find the derivative of $sin(x$)^2 * e^x