# Net velocity of thrown object-lost

1. Jan 22, 2006

### kajalove

hey

I'm currently learning about velocities of objects thrown horizontally.The main point is that initial horizontal speed v[0] will remain constant while vertical velocity will accelerate with a = 10m / s^2.

But when learning formulas there are few things that got me confused:

1)We can calculate net velocity of falling object moment before it hits the ground with:

h --> height

v^2 = v[0]^2 + g^2*t^2 = v[0]^2 + 2*g*h

We got g^2*t^2 = 2*g*h since t^2 = 2h / g

Do we use this formula for net velocity also when an object is thrown in vertical direction with initial speed of v[0]? If that is the case then at least from my understanding this formula doesn't make much sense:

*Firstly, t^2 = 2h / g is true only when object is dropped (and its initial velocity v[0] is 0 => v^2 = v[0]^2 + 2*g*h = 2*g*h)

*If we toss objects vertically with initial speed of v[0],then its final net velocity should be :

v = v[0] + g*t,and not v=sqrt( v[0]^2 + g^2*t^2 )

2)And if formula for net velocity v[0]^2 + g^2*t^2 is indeed the same no matter if object is tossed with v[0] horizontally or vertically from same height, then it would suggest net velocity in both cases is the same. Which again doesn't make much sense to me!

2. Jan 22, 2006

### HallsofIvy

Staff Emeritus
Neither one is "net velocity". They are "vertical component" of the velocity vector. It is, in fact, true that the vertical component of velocity is independent of the horizontal component.

3. Jan 22, 2006

### kajalove

I'm sorry but your reply was way too scarce to help me with any of my questions

4. Jan 22, 2006

### Pyrrhus

What is with this net velocity? You mean final speed (you're using scalars equations).

What type of movement you're describing? Projectile Motion? well it doesn't matter in any case most movements in a rectangular system can be solved by parametric equations, where the parameter is time (t).

5. Jan 22, 2006

### kajalove

Yes,I meant final speed.

I didn't see the need to use vector equations

when object is thrown horizontally,final net velocity is vector object has moment before object hits the ground,and this vector is tangent to the path of object.
When I'm talking about object being thrown vertically(meaning straight down)
with speed V[0],I use word net velocity to describe vertical velocity.

6. Jan 22, 2006

### Staff: Mentor

The equation $v^2 = v_0^2 + 2 g h$ is true regardless of the direction of that initial speed $v_0$; in general, $v^2 \ne v_0^2 + g^2 t^2$. (The first equation is not derived from the latter, since the latter is only true under certain conditions.)
Right, that equation, $v^2 = v_0^2 + g^2 t^2$, happens to be true when the object is thrown horizontally; it's not true in general.

Right, but it's still true that $v^2 = v_0^2 + 2 g h$.

But that's true! (Not that formula, but the fact that final speed of an object that falls a distance h is the same whether it's thrown vertically or horizontally.) The speed is the same, but not the direction.

The key to understanding this is:
(1) Understand the kinematics of accelerated motion and how the equations are derived and interconnect with each other.
(2) Realize that vertical and horizontal motions are independent, and that only the vertical is accelerated.

7. Jan 22, 2006

### kajalove

That I know

But how do we derive formula for net velocity v = sqrt( v[0]^2 + 2*g*h ) when object is thrown from certain height under certain angle(that angle being above or below horizon)?In other words,when besides initial horizontal speed there is also initial vertical speed?

Can you also reason with me in words(not formulas) why final speed of an object that falls distance h will be the same no matter in which direction object is thrown with initial speed v[0]?

8. Jan 23, 2006

### Staff: Mentor

Let's derive the equation $v^2 = v_0^2 + 2 g h$ for the case where $v_0$ represents an initial vertical speed. (Since vertical and horizontal directions are independent, there's no need to worry about an initial horizontal speed: $v^2 = v_x^2 + v_y^2$.)

$$v = v_0 + a t = v_0 -g t$$

Also, distance = average speed x time gives us:
$$-h = \frac{(v + v_0)}{2} t$$
(note the minus sign since it falls a distance h below the starting point)

or:
$$t = -2h/(v + v_0)$$

Plugging that into the first equation, we get:
$$v = v_0 + a t = v_0 + 2g h/(v + v_0)$$

Multiply both sides by $(v + v_0)$ and simplify to get:
$$v^2 = v_0^2 + 2 g h$$

The simplest way to view this is using the concept of energy (which you may not have covered yet). The object has an initial kinetic energy that is independent of its direction of motion. The final kinetic energy is the original plus the work done by gravity. That work depends only on the distance fallen, thus the final kinetic energy (and thus speed) is the same regardless of the direction that the object is initially thrown.

9. Jan 23, 2006

### kajalove

Off topic but I feel it's an important subject:

I'm not shure I understand why h is negative and you still get the same correct result. Is it because you also made acceleration vector g negative( $$v = v_0 + a t = v_0 -g t$$ )?
I know it's off topic, but when you decide to give some variables negative prefix, do you have to be consistent by also making certain other variables negative in order to derive correct result?

Can you tell me some rules I can follow when doing that?
And besides ,why is it important to mark h negative? So we know the direction of an object is downwards?

Back to the subject at hand:

But what if besides having initial vertical speed it would also have initial horizontal speed $$v_h$$? Would formula then be

$$v^2 = v_h^2 + v_0^2 + 2 g h$$ ?

I imagine not since from it we can't derive $$v^2 = v_0^2 + 2 g h$$ where $$v_0^2$$ would represent initial speed (initial vertical and initial horizontal combined). Yup,I'm lost :(

No, I haven't covered it yet. Is there some other way to describe it?

10. Jan 23, 2006

### Staff: Mentor

That's right. (h is negative because it stands for a distance in the negative direction.)
Absolutely; you need a consistent sign convention. (It's not really off topic, since it's required to properly use the kinematic equations.)

It's simple: up is positive; down is negative. Using that sign convention, the acceleration is -g, and the final position of the object is h meters below the starting point thus is -h.

I think you have the right idea, if by $v_0$ you mean the vertical component of the initial velocity. You'd be better off writing it as:
$$v^2 = v_{0,h}^2 + v_{0,v}^2 + 2 g h$$

I don't think you're as lost as you think. First, realize that any initial horizontal component of velocity doesn't change. Second, realize that you get the total velocity (what you are calling "net") this way: $v^2 = v_v^2 + v_h^2$.

So, we've proven, as I think you understand, that:
$$v_v^2 = v_{0,v}^2 + 2 g h$$

And you know that:
$$v_h^2 = v_{0,h}^2$$

Add them up to find the "net" final velocity after falling a distance h:
$$v^2 = v_h^2 + v_v^2 = v_{0,h}^2 + v_{0,v}^2 + 2 g h$$

Which gives us:
$$v^2 = v_0^2 + 2 g h$$
where $v_0^2$ is the total initial speed squared, regardless of direction (since we added the components to get the total).

Got it?

I know of no clever words that would help. (We could describe the kinematic equations in words, but that's the hard way.) I think you are perfectly capable of understanding the above argument if you go over it a few times.

11. Jan 23, 2006

### Staff: Mentor

That's right. (h is negative because it stands for a distance in the negative direction.)
Absolutely; you need a consistent sign convention. (It's not really off topic, since it's required to properly use the kinematic equations.)

It's simple: up is positive; down is negative. Using that sign convention, the acceleration is -g, and the final position of the object is h meters below the starting point thus is -h.

I think you have the right idea, if by $v_0$ you mean the vertical component of the initial velocity. You'd be better off writing it as:
$$v^2 = v_{0,h}^2 + v_{0,v}^2 + 2 g h$$

I don't think you're as lost as you think. First, realize that any initial horizontal component of velocity doesn't change. Second, realize that you get the total velocity (what you are calling "net") this way: $v^2 = v_v^2 + v_h^2$.

So, we've proven, as I think you understand, that:
$$v_v^2 = v_{0,v}^2 + 2 g h$$

And you know that:
$$v_h^2 = v_{0,h}^2$$

Add them up to find the "net" final velocity after falling a distance h:
$$v^2 = v_h^2 + v_v^2 = v_{0,h}^2 + v_{0,v}^2 + 2 g h$$

Which gives us:
$$v^2 = v_0^2 + 2 g h$$
where $v_0^2$ is the total ("net") initial speed squared, regardless of direction (since we added the components to get the total).

Got it?

I know of no clever words that would help. (We could describe the kinematic equations in words, but that's the hard way.) I think you are perfectly capable of understanding the above argument if you go over it a few times.