Vertical movement: velocity / time graph, height estimation

[moenste]

Homework Statement

A ball is thrown vertically upwards and caught by the thrower on its return. Sketch a graph of velocity (taking the upward direction as positive) against time for he whole of its motion, neglecting air resistance. How, from such a graph, would you obtain an estimate of the height reached by the ball?

2. The attempt at a solution
I think the graph should be:

Where time is on the X-axis and velocity on the Y-axis. And maximum height: v² = u² + 2as with v = 0. (v = velocity at t, u = velocity at t = 0, s = distance travelled)

But I am not sure about this.

 

[brainpushups]

Do you know the algebraic expression for velocity as a function of time for an object in free fall? How does this equation compare with your graph?

 

[moenste]

Do you know the algebraic expression for velocity as a function of time for an object in free fall? How does this equation compare with your graph?

So my graph is correct and the formula should be: s = vt ?

 

[brainpushups]

So my graph is correct and the formula should be: s = vt ?

I didn't say that. What type of equation is s = v*t (linear, quadratic, inverse, logarithmic, etc.) and what type of graph have you drawn?

 

[moenste]

I didn't say that. What type of equation is s = v*t (linear, quadratic, inverse, logarithmic, etc.) and what type of graph have you drawn?

The graph is quadratic and the formula is linear (not sure though).

 

[brainpushups]

The graph is quadratic and the formula is linear (not sure though).

Yup. So the two contradict. The graph cannot be the same as the equation. For starters your graph shows that the initial velocity of a thrown object would be zero. That can't be right.

 

[moenste]

Yup. So the two contradict. The graph cannot be the same as the equation. For starters your graph shows that the initial velocity of a thrown object would be zero. That can't be right.

Shouldn't the initial velocity at t = 0 be also zero? t = 0, the moment when the ball is in the hand.

But if we take the stating point when the ball is already thrown, then, as I understand, the velocity should be zero at the maximum height and should be negative after the max. height (because it is falling). So, I would graph it like this:

Where X is time, Y is velocity. We start at (0, 3), when t = 0 and v = 3. At t ~ 4.5, v = 0 (the max. height point) and then the velocity starts to increase, but because the ball is falling, is it negative. The line which is left from (0, 3) is not included. But then I don't know how to obtain an estimate of the height reached by the ball. If velocity is zero at maximum height, then s = vt cannot be used.

 

[brainpushups]

Shouldn't the initial velocity at t = 0 be also zero? t = 0, the moment when the ball is in the hand.

Questions like this refer to the initial velocity of the ball as the velocity that the ball has when it separates from the hand.

Your graph looks good.

Where X is time, Y is velocity. We start at (0, 3), when t = 0 and v = 3. At t ~ 4.5, v = 0 (the max. height point) and then the velocity starts to increase, but because the ball is falling, is it negative. The line which is left from (0, 3) is not included. But then I don't know how to obtain an estimate of the height reached by the ball. If velocity is zero at maximum height, then s = vt cannot be used.

Have you learned about the relations of area under a curve for the graphs of motion (area of acceleration vs. time, and velocity vs. time)? What are the units of the area bounded by the curve in your sketch?

 

[moenste]

Have you learned about the relations of area under a curve for the graphs of motion (area of acceleration vs. time, and velocity vs. time)? What are the units of the area bounded by the curve in your sketch?

https://www.khanacademy.org/science...why-distance-is-area-under-velocity-time-line
As I understand, this is what you are suggesting. The area under the triangle is the distance traveled (in my case height).

The triangle which is between (0, 3) and (4.5, 0) is what I should examine to find the height. And area of a triangle = 1/2 * base * height = 1/2 * 4.5 * 3 = 6.75 m. So the ball achieved a height of 6.75. And in case I had a question "what's the distance traveled by the ball" then it's just height * 2 (because the ball returns in the hand): 6.75 * 2 = 13.5 meters.

Wouldn't it be also correct to: s = vt so s = 3 m/s * 4.5 s = 13.5 meters / 2 = 6.75 m, because we need only the half of the area.

Right?

One more question: e.g. we have initial velocity 20 m/s and time to max height is 5 seconds. So we draw: (0, 20) as first point, (5, 0) as max height and (10, -20) as the moment when the ball returns to the hand, end of line. And this would be correct, right? As you say "Questions like this refer to the initial velocity of the ball as the velocity that the ball has when it separates from the hand." so it means the graph starts from 20 and ends and -20 and it's correct (velocity should not be equal to zero at the moment when the hand throws the ball and when it cathes it, this moment is not shown on the graph, as I see it). Did I understand this point correct?

Thank you once again :).

 

[brainpushups]

The triangle which is between (0, 3) and (4.5, 0) is what I should examine to find the height. And area of a triangle = 1/2 * base * height = 1/2 * 4.5 * 3 = 6.75 m. So the ball achieved a height of 6.75. And in case I had a question "what's the distance traveled by the ball" then it's just height * 2 (because the ball returns in the hand): 6.75 * 2 = 13.5 meters.

Indeed. One thing to point out is that the area is sign sensitive. The total area for the ball's trajectory will include equal areas above and below the x axis. The total area gives the displacement (zero) of the ball. The distance traveled ignores the sign of the area and this is why the total area is twice that of the triangle for height.

Wouldn't it be also correct to: s = vt so s = 3 m/s * 4.5 s = 13.5 meters / 2 = 6.75 m, because we need only the half of the area.

The equation you have used is only valid for uniform motion or when the average velocity (or speed in the case of distance) is used. Because the average speed is exactly half of the initial/final speed it does work here, but I would be careful applying the uniform motion equation to motion which is not uniform.

One more question: e.g. we have initial velocity 20 m/s and time to max height is 5 seconds. So we draw: (0, 20) as first point, (5, 0) as max height and (10, -20) as the moment when the ball returns to the hand, end of line. And this would be correct, right?

Yes.

As you say "Questions like this refer to the initial velocity of the ball as the velocity that the ball has when it separates from the hand." so it means the graph starts from 20 and ends and -20 and it's correct (velocity should not be equal to zero at the moment when the hand throws the ball and when it cathes it, this moment is not shown on the graph, as I see it). Did I understand this point correct?

Yes.

 

[moenste]

...

Thank you!