Why is the velocity-time graph for vertical projectile motion linear and not parabolic?

[polyperson]

If an object is thrown vertically upwards at an initial velocity of n ms-1 (metres per second) and returns to the point where it began, the final velocity will be n ms-1 and the velocity at the highest point reached will be 0 ms-1.

But...what would be the shape of the velocity-time plot? Would it be an upside down parabola or would the acceleration/retardation be uniform (surely it would because g=9.81 ms-2) giving an upside down V-shape?

I think it's an easy question but I can't quite decide which is true!

 

[Chi Meson]

Can you say what the acceleration is throughout the flight? Is the acceleration constant?

Looking at a v-t graph, what tells you the acceleration?

 

[polyperson]

Acceleration is the slope of the plot at all points and I'm sure the acceleration in this case must be constant i.e. 9.81 ms-2. That tells me it must be an upside down V-shape rather than parabolic. Have I answered my own question? The difficulty for me is that I find it hard to grasp the idea that the object accelerates to n ms-1 the instant it begins the journey i.e u = n ms-1. If it was at rest, surely n = 0 ms-1 and accelerattion is not constant!

 

[Chi Meson]

The acceleration is the slope of a v-t graph, and the acceleration is constant. You've got that, so why does your slope change? If acceleration is constant, then slope stays constant. (NOT a "V" shape).

 

[polyperson]

That's really thrown me. The velocity at t=0 seconds is, say, 100ms-1. As the object reaches maximum height, v = 0 ms-1. As it begins to fall, it accelerates back up to 100 ms-1 at the point where it was released i.e. it's an upside down V-shape with the slope either side symmetrical (g or -g).

If it had no slope, the velocity would be constant which it very definitely isn't.

The real problem to me is the way it is framed. u must be 0 ms-1 not some other value, surely?

 

[Chi Meson]

It has a slope, and the slope is constant.

The velocity is not exactly the same at the end, is it? Which way is velocity "pointing" at start and finish? Is one way considered "negative"?

What would a negative velocity look like on a v-t graph?

 

[polyperson]

Wait, I've got it. Thanks for your hints!

It's a straight line with negative slope because g = constant = -9.81 ms-1 and the velocity after the object reaches the highest point becomes negative.

Thankyou Chi

 

[Chi Meson]

Touche, mon frere.