Very Vague Graphing Motion Problem

[rakeru]

Homework Statement

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, beginning the instant the stone leaves the thrower’s hand. Construct the corresponding motion graphs taking the acceleration due to gravity as exactly 10m/s² . Ignore air resistance. In all three motion graphs, the unit of time is in seconds and the unit of displacement is in meters. In plotting the points, round-off the coordinate values to the nearest integer.

I need to make three graphs with the vertical displacement, vertical velocity, and vertical acceleration. The graphs are from t=0 to t=6 s.

I attached a screenshot of the diagram they gave me.

Homework Equations

V=V₀+at

y=y₀+V₀t+0.5at²

The Attempt at a Solution

I need to find the initial velocity! It doesn't tell me. I'm not sure if I can find the initial velocity from what is given. The image shows that the highest point is at t=2s. But does that mean that v=0 at that point?? There seems to be two points up there.. It's confusing me.

Can I use V = V₀+at to find it?
I already tried doing that and I got 20 m/s from (at the highest point):

0=V₀+(-10m/s²)(2 s)

Another thing! Is the gravitational acceleration negative? The question didn't tell me which direction is positive! How am I supposed to know??

So is V₀= 20m/s??

 

[SteamKing]

In your experience, in which direction does gravity act? Do things go flying into the air for no particular reason?

The problem statement is very clear. A stone is thrown upward from the edge of a cliff. Can you find that on the attached diagram? The stone reaches its max. height 2 s after it is thrown. What happens at max. height?

The diagram is drawn the way it is, I think, for clarity, to show that the stone changes direction 2 s after it is thrown from the cliff.

Stay calm and work through the problem. There are no tricks here.

 

[haruspex]

Is the gravitational acceleration negative? The question didn't tell me which direction is positive! How am I supposed to know??
So is V₀= 20m/s??

Looks right.
It's up to you to choose whether up or down is positive. It doesn't matter as long as you are consistent. Usual is to take up as positive, and that seems to be what you have done.

 

[rakeru]

Thanks so much. I was just unsure.

 

[Nickie]

First of all, it is important to note that the question states to ignore air resistance, so we can assume that the only force acting on the stone is gravity. This means that the gravitational acceleration will always be pointing downwards towards the ground, which we can take as the negative direction in our motion graphs.

Now, to find the initial velocity (V0), we can use the fact that the stone reaches its maximum height at t=2s. This means that at this point, the vertical velocity (Vy) is equal to 0. We can use the equation V=V0+at to solve for V0:

0 = V0 + (-10m/s^2)(2s)

V0 = 20 m/s

This confirms your initial calculation, so your answer of V0=20m/s is correct.

Next, we can plot the motion graphs using this initial velocity and the given time frame of t=0 to t=6s. For the vertical displacement graph, we can use the equation y=y0+V0t+0.5at^2. Since the stone starts at a height of 0 (we can take this as our reference point for displacement), the equation simplifies to y = 0 + (20m/s)t + 0.5(-10m/s^2)t^2. We can then plot this equation on a graph, with time (t) on the x-axis and displacement (y) on the y-axis.

For the vertical velocity graph, we can use the equation V=V0+at. Again, since the stone starts with an initial velocity of 20m/s, the equation becomes V = 20m/s + (-10m/s^2)t. We can plot this equation on a graph, with time (t) on the x-axis and velocity (V) on the y-axis.

Lastly, for the vertical acceleration graph, we can simply plot a horizontal line at -10m/s^2, as this is the constant acceleration due to gravity acting on the stone.

In summary, to solve this problem and construct the motion graphs, we used the equations V=V0+at and y=y0+V0t+0.5at^2, along with the given information of a maximum height at t=2s, to find the initial velocity and plot the motion graphs. It is important to pay attention to the details given in the question and