Consistency of premises in a reasoning regarding ( projectile) motion:on using( conterfactually)the inertia principle to determine the upward velocity

Question

My question deals mainly with the reasoning used to solve the problem.

Suppose I know a projectile is moving in the XY plan and is sent at the origin with an initial velocity $v_i$ the vector of which making a positive angle $\alpha$ with the horizontal line ( the Ox axis).

Suppose the question is : what is the vertical velocity at time $t$ ?

Can I reason in the following way?

(1) If the projectile were submitted to no force, then ( by the principle if inertia) it would keep its initial upward velocity : $v_i sin\alpha$ , for all $t$.

(2) But the projectile is actually submitted to the gravitational force of the earth, and therefore is accelerated at the rate $g \space ( = 9.81m/s^{-2})$. So P's downward velocity at $t$ is $-gt$.

(3) Therefore, P's total vertical velocty at $t$ is : $v_i t + (- gt)$.

It seems to me that the conclusion rests on 2 premisses corresponding to 2 different scenarios ( no resultant force for premise (1) and gravitational force for premise (2)); and these scenarios are not consistent.

How can premise (1) be true if (2) is also true?

Answer 1

You are correct when you say that the velocity won't change when the body has been subjected to no force.But the second case will be little different : When the body is acted upon by gravity the force due to gravity acts vertically downward .So the vertical component of the velocity will change but horizontal component will not change since there is no force in the horizontal direction .So ,

initial velocity = $v$;

vertical component of velocity = $v \sin \alpha$

horizontal component of velocity = $v \cos \alpha$

gravity acts in vertical direction so it has no component in horizontal direction(since $g.\cos \frac \pi2 = 0$) and won't affect horizontal velocity.

So only vertical velocity is altered. To find velocity after time 't' :

We know,$$ v = u + at$$ $$v_{final} = v\sin \alpha + (-g)t$$

So final vertical component of velocity = $v_{final}$; and,

final horizontal component of velocity = $v \cos \alpha$

So net velocity at time 't' = $\sqrt{v_{final}^2 + (v\cos \alpha)^2}$

So that's the math behind it

Answer 2

Can I reason in the following way?

As you mentioned, the reasoning is flawed because it relies on two premises which are not only counterfactual but mutually contradictory. However, the math works out for a different reason:

The equation that describes a projectile is $\vec r'' = \vec g$ where $\vec r$ is the position vector at any time, $t$, and $\vec g$ is the gravitational field vector (pointing down).

Now, notice that the projectile equation has the form of a second-order non-homogenous linear differential equation. Non-homogenous differential equations have the feature that their general solution can be written as the sum of any particular solution to the non-homogenous equation and the general solution of the complementary equation. In this case the complementary equation is $\vec r''=0$.

Despite the reasoning being flawed, it works because of this property of non-homogenous linear differential equations. Your statement (1) is solving the complementary equation $\vec r''=0$. Your statement (2) is a particular solution of the non-homogenous equation. So the overall solution is indeed their sum.

This is, in fact, a standard strategy for solving non-homogenous differential equations. I would not rely on this physical reasoning, but it is perfectly justified to set up the equations based on sound physical reasoning and then solve them as you solved it based on this mathematical justification.

Answer 3

Either premise 1 or premise 2 will be true, but not at the same time. The projectile is either launched in a gravitational field or it is not. It cannot be both ways at once so premise 1 and 2 do not conflict with each other.