Motion Diagram and Gravity Graphing Question

[DefaultName]

A stone is thrown upward from the edge of a cliff, reaches its maximum height, and then falls down into the valley below. A motion diagram for this situation is given View Figure https://www.physicsforums.com/attachment.php?attachmentid=6246&stc=1&d=1139278282 , beginning the instant the stone leaves the thrower’s hand. Construct the corresponding motion graphs. Ignore air resistance. In all three motion graphs, one unit of time is equivalent to one unit of time (one second) in the given motion diagram.

Construct a graph corresponding to the stone's vertical displacement, y(t).

Ok, I know that the intial position is 0. However, when I want to find the displacement value at t = 2... I do this:

I use Vf = Vo + at to find the intial velocity thrown above.. I use -9.8 for a, t for 2, vf for 0 since it's at the top.

Then I plug the Vo into Vf^2 = Vo^2 + 2as, and solve for s.

However, I get 19.6 and that isn't within bounds for me to draw the graph..

What am I doing wrong? or Could I use another technique?

Thanks

 

[Astronuc]

Uniform acceleration, -9.8 m/s², i.e. acceleration is downward.

Two equations:

y(t) = y_o + 1/2 a t² gives y(t), neglecting wind resistance.

v(t) = v_o + at

 

[quicknote]

Hello,

Thank you for sharing your question and thought process. It seems like you are on the right track in terms of understanding the concepts involved in constructing a motion graph for this scenario.

To answer your question, the issue may lie in the units you are using. In the given motion diagram, one unit of time is equivalent to one second. However, you are using the value of -9.8 for acceleration, which is in units of meters per second squared (m/s^2). To match the units of time, you should use -9.8 m/s^2 for acceleration.

Also, when using the equation Vf = Vo + at, it is important to consider the direction of the acceleration. In this case, the stone is moving upward, so the acceleration should be positive. Therefore, the equation would be Vf = Vo + 9.8t. This would result in a final velocity of 19.6 m/s at t = 2 seconds.

Using this value in the equation Vf^2 = Vo^2 + 2as, we can solve for s and get a displacement value of 19.6 meters. This is within the bounds of the given motion diagram, so you should be able to plot it on your graph.

In terms of using another technique, you could also use the equation y = yo + vot + 1/2at^2 to find the displacement at a specific time. In this case, yo (initial position) and vo (initial velocity) would both be 0, and the equation would simplify to y = 1/2at^2. Plugging in the values of a = 9.8 m/s^2 and t = 2 seconds would also result in a displacement of 19.6 meters.

I hope this helps clarify the issue and provides you with some additional techniques for constructing motion graphs. Keep up the good work!