1
$\begingroup$

I am a High School Student and I've started to learn about Projectile Motion. One of the Assumptions made in the text is that the horizontal acceleration must be $0$ for the Equations of Projectile Motion to be applicable. I'm wondering what the case would be if the Horizontal Acceleration was non-zero and more generally a variable function of time $t$. I'm guessing the Trajectory won't be Parabolic then. How can I give Mathematical Relevance to my thought?

Thanks in Anticipation.

Improve this question
$\endgroup$
2
  • $\begingroup$ Hi, and welcome to stack exchange! What have you tried so far? $\endgroup$
    – joshuaronis
    Dec 29, 2018 at 15:16
  • $\begingroup$ I've tried using the following information:$$\begin{pmatrix} u_x & a_x\\u_y & a_y \end{pmatrix}=\begin{pmatrix} u \cos \theta & a\\u \sin \theta & -g \end{pmatrix}$$For the Horizontal Motion I got the following Equations of Motion:$$v_x=u \cos \theta+at \cdots (1)$$ $$x=ut \ cos \theta +\frac{1}{2}at^2 \cdots (2)$$For the Vertical Motion I got: $$v_y=u \sin \theta -gt\cdots (3)$$ $$y=ut \sin \theta -\frac{1}{2}gt^2 \cdots (4)$$From Equation $(2)$ we get $t$ in terms of $x$ as $t=\frac{-u \cos \theta}{a} \pm \frac{\sqrt{D}}{2a}$ I can't figure which $t$ to use for substitution into $(4)$ $\endgroup$
    – Paras Khosla
    Dec 30, 2018 at 7:25

2 Answers 2

Reset to default
0
$\begingroup$

Well, if all you want is to show it as a mathematical function, why don't you just leave it as a parametric function of time?

https://www.desmos.com/calculator/idydulmnoj

Improve this answer
$\endgroup$
0
$\begingroup$

First, the mathematical relevance would be writing down Newton's second law in vector component form. The "assumption" you mention is quite real, namely that near the surface of the earth the force of gravity acting on us is approximately constant and points down towards the surface.

Newton's second law of motion would be:

-mg = ma_y

0 = m*a_x

where y direction is vertical and x horizontal, notice that mass cancels out.

For your question Newton's second law would give...

a_y = -g

a_x = some constant

or

ax = f(t), some function of time.

Acceleration is the second derivative of position so either of these would give you a new equation to find x(t) and y(t). Now, since you are in high school physics I'd ask if you've have pre-calculus or calculus 1 yet. If not then going much further may not help you see how things change but I will add a little more.

What you get is a differential equation and there are techniques for solving these. You may be at the level where you have not yet learnt Newton's laws of motion but are just learning Kinematics. That's okay, we can start from that point.

a_x = dv_x/dt = d/dt(dx/dt)

a_y = dv_y/dt = d/dt(dy/dt)

These are the equations that relate acceleration to velocity and position. The kinematics you learnt came from these equations with a_x = 0 and a_y = -g.

For your first question, what if there is a constant acceleration in the x direction, this will in fact produce a parabola but the symmetry axis will not be the vertical axis but the axis defined by the acceleration vector (a_x, a_y).

For a time dependent acceleration you need to "integrate" the second derivative twice and find the correct x(t), and y(t). That is a basic procedure.

In the real world we have air resistance that depends on the velocity of the projectile, the force is proportional to (-v_x, -v_y). The equations of motion with "linear" air resistance are solvable and you get a correction to the parabola path that involves exponential functions. You can look this up on Wikipedia, search projectiles with air resistance.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.