# Why does drop time remain constant regardless of initial horizontal velocity?

1. Nov 20, 2008

### charlatain

So in projectile motion, an object launched horizontally traveling over a certain displacement (Δx), fall time remains constant regardless of what this horizontal velocity was.

I understand that in this equation:

Δx = v(avg) × Δt

as horizontal velocity increases, displacement also increases, with time serving as the constant of variation. Mathematically, this makes sense. But in reasoning, I'm having a hard time grasping the concept. Just why does time remain constant?​

2. Nov 20, 2008

### aerospaceut10

You should think of the velocity of the object in two components, the vertical and horizontal. Because the time for the object to hit the ground from that height is not going to change since the height hasn't been changed...you only are concerned with the initial height, gravity, and the initial velocity. All in the vertical direction, that is.

3. Nov 20, 2008

### montoyas7940

Initial vertical velocity, acceleration due to gravity and vertical displacement are not changed by horizontal velocity (at a small scale). There is nothing else to change so why expect time to change. It does seem counter-intuitive but the math reveals the truth.

4. Nov 20, 2008

### uart

Yes it does so if you ignore air resistance and any other aerodynamic forces.

Why? Well it should be pretty obvious that if you exclude any aerodynamic forces then the horizontal component of velocity can have no effect at all on the vertical forces (and visa versa). Does that make sense?

5. Nov 20, 2008

### charlatain

I suppose.

So what I understand (of course, air resistance being negligible):

The horizontal (x) motion of an object is entirely independent on its vertical (y) motion. Initial vertical velocity, acceleration due to gravity and vertical displacement are not affected by horizontal velocity, therefore time is not expected to change either. For example, simply dropping an object down vertically or launching it horizontally at any velocity will yield the same drop time simply because horizontal and vertical components CANNOT influence one another, given that the height, initial vertical velocity, acceleration due to gravity are kept constant.

Is that somewhere near?

6. Nov 22, 2008

### mordechai9

You are on the right track here. Allow me to be very clear, and systematic.

Essentially we are talking about Newton's 2nd law, F=ma. The key concept is that this equation is a vector equation. It decomposes into three different equations, relating the forces in the x-direction to the acceleration in the x-direction, and similarly for the y and z directions. Let's call these the x-force, y-force, and z-force equations.

Consider the y-force equation. The gravity force, as we know, always points downwards; hence the gravity force shows up in the y-force equation. But the gravity force is the only force pointing in the y-direction; so the the acceleration in the y-direction is only governed by the gravity force. Hence the "drop time" as you call it has no dependency on the velocity in the x-direction.

I believe this is the most systematic reasoning by which you would come to your conclusion. Please reply if you still don't understand.