# Projectile motion problems with No horizontal acceleration

• opus
In summary: Va and Vb are moving in opposite directions.In summary, the conversation discusses the concept of projectile motion and the relationship between horizontal and vertical components. It is explained that in these types of problems, the horizontal component of the ball's path has no acceleration and therefore its initial velocity remains constant. The speaker also mentions that the concept of "starting" a problem can be subjective and depends on one's perspective.
opus
Gold Member
For projectile motion problems, we say that the horizontal component of the ball's path has no acceleration. The vertical component does, which is due to gravity, but in projectile motion it is true that horizontal and vertical motion are independent of each other.

My question is in regards to the lack of acceleration for the horizontal component. We initially start with a ##v_0=0 m/s##. A tiny tiny amount of time after the ball is hit, it must have ##v=0.1m/s##. A tiny amount of time after that, it must have ##v=0.2 m/s## this will continue until it reaches its maximum velocity which it will hold until it is no longer in flight (neglecting air resistance). So is it not true, that the horizontal component of the ball's flight does have acceleration? This logic does seem to present a problem to me though because if there is nothing acting against it, it would never hit a maximum velocity and would continually increase speed.
So what is wrong with my reasoning?

opus said:
My question is in regards to the lack of acceleration for the horizontal component. We initially start with a ##v_0=0 m/s##. A tiny tiny amount of time after the ball is hit, it must have ##v=0.1m/s##.
No. Since there is no horizontal acceleration if the horizontal ##v_0=0## then it will continue to be 0 indefinitely. The horizontal velocity would never become 0.1 m/s. In general for these problems, whatever the initial horizontal velocity is what it will remain.

opus
But the velocity has changed from 0 to a non-zero value, which implies a rate of change in velocity which is the definition of acceleration. Also, how could the horizontal velocity never become 0.1 m/s if it has to go from 0 to a new value?

opus said:
But the velocity has changed from 0 to a non-zero value,
Not if the acceleration is 0. Which is a given in these problems. It stays at the initial velocity, whatever that is.

opus
Dale said:
Not if the acceleration is 0. Which is a given in these problems. It stays at the initial velocity, whatever that is.
How has the velocity not changed from a zero value if it's moving, considering velocity is defined as the rate at which the position changes over a given time interval. The position of the projectile is changing, so there must be velocity right?

The velocity in the horizontal component is fixed for these problems.

Try to find Alonso and Finn : Fundamental University Physics, at the school library. It goes over these derivations and walks you through them. Then it generalizes.

Or purchase the book. A good investment.

So do we say that problem doesn't "start" until the ball moves, in which case the initial velocity is not zero?

Remember. That velocity for these types of problems has 2 components, the x and y component. Just because the velocity in the x direction is 0, it does not mean that velocity while the object has been in motion is 0. The force of gravity is acting on the object.

MidgetDwarf said:
Just because the velocity in the x direction is 0
This is where I don't understand. How can the horizontal velocity be zero? It's changing position over the x- axis which is literally the definition of velocity.

opus said:
How has the velocity not changed from a zero value if it's moving,
If it is moving then it started out moving. In these problems there is no horizontal acceleration. So if it starts at v=0 then it stays at v=0. If it starts at v=10 then it stays at v=10. It never changes from a 0 velocity to a non-zero velocity.

You seem to believe that it must start at v=0. This is not correct

opus
Hmm I must have missed something key at the beginning. Let me go back and re read the section.

The object having a zero acceleration along the x-axis does not mean it must have a zero velocity along the x axis.

It simply means its x velocity - whatever it may be at the start - stays constant.

opus
I think that this is mostly a communication problem.
opus said:
So do we say that problem doesn't "start" until the ball moves,
Yes. These problems are often referred to as “ballistic projectile” problems. As long as the object in question is being accelerated (while the ball is in contact with the bat, or the cannon is traveling down the barrel of the gun), it is not yet considered a “projectile”, and certainly not a ballistic one.
However, to say…
in which case the initial velocity is not zero?
…is entirely a matter of perspective. From one viewpoint Va, a golf ball sitting on a T has zero velocity. From another viewpoint Vb, that same ball has zero velocity after it has been hit by the driver and is in flight. From Va, the ball has some positive forward velocity after it has been hit, from Vb, the ball had some negative forward velocity until after the golf club game along and “stopped” it. Either way, the result is the same; a T’d off golfball.

So if the problem begins by saying that the projectile in flight has zero velocity, that is simply setting the perspective from which they want you to proceed.

opus
LURCH said:
As long as the object in question is being accelerated (while the ball is in contact with the bat, or the cannon is traveling down the barrel of the gun), it is not yet considered a “projectile”, and certainly not a ballistic one.
Ok this was my confusion. So we call it a projectile motion once whatever got it moving has stopped getting it moving. That is, once it's actually in flight.

DaveC426913
opus said:
Ok this was my confusion. So we call it a projectile motion once whatever got it moving has stopped getting it moving. That is, once it's actually in flight.
Correct! Now, we celebrate with herring!

opus
opus said:
Ok this was my confusion. So we call it a projectile motion once whatever got it moving has stopped getting it moving. That is, once it's actually in flight.
Yes problems normally say something like . .. "a projectile is launched at some velocity" or "calculate the launch velocity", in which case we aren't usually interested in how it got to that velocity.

That's not to say we couldn't write a problem that required you work out the acceleration during the launch phase but typically that would be a separate calculation.

I've come across problems involving a coin that was slid along a table and falling off the edge. For the first part we had to calculate the effect of friction decelerating the coin so we could work out the velocity with which it left the edge. Then for the second part we assumed there was no horizontal deceleration (Eg zero air resistance) so we could work out where it hit the floor. Even this problem ignored how the coin was accelerated from rest. It just said a man slides a coin at an initial velocity towards the edge...

opus
Awesome thanks all!

## What is projectile motion?

Projectile motion is the motion of an object through the air, under the influence of gravity, after being given an initial velocity. This type of motion is commonly seen in sports such as throwing a ball or shooting a projectile.

## What is the difference between horizontal and vertical acceleration?

Horizontal acceleration refers to the change in velocity of an object in the horizontal direction, while vertical acceleration refers to the change in velocity in the vertical direction. In projectile motion problems with no horizontal acceleration, the object's velocity in the horizontal direction remains constant, while its velocity in the vertical direction is affected by gravity.

## How do you calculate the time of flight for a projectile with no horizontal acceleration?

The time of flight for a projectile with no horizontal acceleration can be calculated using the equation t = 2v₀sinθ/g, where t is the time of flight, v₀ is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity (9.8 m/s²).

## What is the maximum height reached by a projectile with no horizontal acceleration?

The maximum height reached by a projectile with no horizontal acceleration can be calculated using the equation h = (v₀sinθ)²/2g, where h is the maximum height, v₀ is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity (9.8 m/s²).

## How do you determine the range of a projectile with no horizontal acceleration?

The range of a projectile with no horizontal acceleration can be calculated using the equation R = v₀²sin2θ/g, where R is the range, v₀ is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity (9.8 m/s²). Alternatively, the range can also be determined by multiplying the time of flight by the horizontal velocity (R = v₀cosθ * t).

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