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A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of v0 = 19.0 m/s. The cliff is h = 49.0 m above a flat, horizontal beach as shown in the figure.A student stands on the edge of a cliff with his hand a height h above a flat stretch of ground below the clifftop. The +x-axis extends to the right along the ground and the +y-axis extends up from the ground to the top of the cliff. The origin O of the coordinate plane is directly below the student's hand where the base of the cliff meets the flat ground. The student throws a stone horizontally rightward with initial velocity vector v0. The stone falls with a parabolic trajectory, hitting the ground with a velocity vector v that points down and right. Vector g points straight down.(a) What are the coordinates of the initial position of the stone?x0= my0= m(b) What are the components of the initial velocity?v0x= m/sv0y= m/s(c) Write the equations for the x- 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.)vx= vy= (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.)x= y= (e) How long after being released does the stone strike the beach below the cliff? s(f) With what speed and angle of impact does the stone land?vf= m/s= ° below the horizontal

A Student Stands At The Edge Of A Cliff And Throws A Stone Horizontally Over The Edge With A Speed Of V0 190 Ms The Cliff Is H 490 M Above A Flat Horizontal Bea class=

Sagot :

We will have the following:

a)

[tex]\begin{gathered} x_0=0m \\ \\ y_0=49m \end{gathered}[/tex]

b)

[tex]\begin{gathered} v_{0x}=19m/s \\ \\ v_{0y}=0m/s \end{gathered}[/tex]

c)

[tex]\begin{gathered} v_x=v_{ix}+a_xt\Rightarrow v_x=19m/s+(0m/s^2)t\Rightarrow v_x=19m/s \\ \\ v_y=v_{iy}+a_yt\Rightarrow v_y=0m/s+(9.8m/s^2)t\Rightarrow v_y=(9.8m/s^2)t \end{gathered}[/tex]

d)

[tex]\begin{gathered} x=v_{ix}t+\frac{1}{2}a_xt^2\Rightarrow x=19m/s+\frac{1}{2}(0m/s^2)t^2\Rightarrow x=(19m/s)t \\ \\ y=v_{iy}t+\frac{1}{2}a_yt^2\Rightarrow y=(0m/s)t+\frac{1}{2}(9.8m/s^2)t^2\Rightarrow y=\frac{1}{2}(9.8m/s^2)t^2 \end{gathered}[/tex]

e)

[tex]\begin{gathered} 49m=\frac{1}{2}(9.8m/s^2)t^2\Rightarrow t=\sqrt{\frac{49m}{4.9m/s^2}} \\ \\ \Rightarrow t=\sqrt{10}s\Rightarrow t\approx3.2s \end{gathered}[/tex]

So, the time it takes is sqrt(10) s, that is approximately 3.2s.

f)

[tex]\begin{gathered} v_f=\sqrt{(19m/s)^2+((9.8m/s^2)(\sqrt{10}s))^2}\Rightarrow v_f=36.35106601...m/s \\ \\ \Rightarrow v_f\approx36.4m/s \end{gathered}[/tex]

So, the final velocity is approximately 36.4 m/s.

[tex]\begin{gathered} 19m/s=(36.4m/s)cos(\theta)\Rightarrow cos(\theta)=\frac{19m/s}{36.4m/s} \\ \\ \Rightarrow\theta=cos^{-1}(\frac{19}{36.4})\Rightarrow\theta=58.53497293... \\ \\ \Rightarrow\theta\approx58.5 \end{gathered}[/tex]

So, the angle would be approximately 58.5°.

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