Why is V_x of a projectile constant?

[NooDota]

Homework Statement

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?

Homework Equations

V_x = V*Cos(theta)

The Attempt at a Solution

 

[rude man]

Homework Statement

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?

Homework Equations

V_x = V*Cos(theta)

The Attempt at a Solution

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

 

[PeroK]

I

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

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

 

[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?

 

[PeroK]

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?

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.

 

[SammyS]

I

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

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.

 

[PeroK]

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

 

[rude man]

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.

Thanks Sammy. Owe you one!

 

[rude man]

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?

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

 

[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?

 

[SammyS]

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?

That's no longer simple projectile motion.

 

[rude man]

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?

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