# Why is V_x of a projectile constant?

1. Jul 2, 2015

### NooDota

1. The problem statement, all variables and given/known data

Why is the horizontal velocity for a projectile constant?

I understand it physically, since the vertical and horizontal velocities aren't related and don't affect each other, g will only affect V_y and neglect V_x, so it's constant.

But maths-wise, isn't V_x = V*Cos(theta)? How can it be constant if the angle is changing?

2. Relevant equations

V_x = V*Cos(theta)

3. The attempt at a solution

2. Jul 2, 2015

### rude man

V is also changing, such that Vcosθ is constant.
Good question!

3. Jul 2, 2015

### PeroK

I
$\theta$ refers to the initial angle at launch, not the changing angle at which the projectile moves thereafter.

Last edited: Jul 2, 2015
4. Jul 2, 2015

### NooDota

But the law for the horizontal velocity is V_x = V*Cos(theta) where theta is the current angle the velocity vector makes with the x+ axis, right?

5. Jul 2, 2015

### PeroK

No, that's wrong. $v$ is the magnitude of the initial velocity and $\theta$ is the initial angle. That's why $v_x$ is constant.

6. Jul 2, 2015

### SammyS

Staff Emeritus
That's only true if θ is defined that way. Clearly OP considers θ to be changing along the trajectory.

In this case we would likely θ0 as the initial value of θ,

As usual, rude man has answered well. - - short and complete.

7. Jul 2, 2015

### PeroK

Apologies, my mistake. I see what the OP was asking.

8. Jul 2, 2015

### rude man

Thanks Sammy. Owe you one!

9. Jul 2, 2015

### rude man

You are correct, and vcosθ is constant unless you try to include air friction. Which you don't want to do, trust me!

10. Jul 2, 2015

### NooDota

Okay, thanks.

Another question, it's not worth making a thread for.

If the velocity of a projectile is given by V = 25i -4.9j (y is directed upwards), has the projectile reached its highest point yet? The book says yes.

Since it's moving downwards, that means yes. But what if I initially threw it down, and it bounced back up higher than its starting point?

11. Jul 2, 2015

### SammyS

Staff Emeritus
That's no longer simple projectile motion.

12. Jul 2, 2015

### rude man

Then its V would no longer be i 25 - j 4.9.