﻿ Two dimensional motion

# Two dimensional motion

Two dimensional motion (2-D) refers to the motion of an object on a curved path on a plane or the motion on two or more straight paths on a plane. An example for a 2-D motion is a projectile motion. In this page, we will focus on projectile motion, and relative velocity in 2 dimensions.

## Projectile motion

In free fall motion, you studied 1-D motion of objects in air. A projectile motion is the 2-D motion of objects in air. The term projectile is used to refer to an object that is launched either horizontally or at an angle from the horizontal into the air and is only under the influence of gravity. The motion of a projectile is called projectile motion. Since any object in air has air resistance acting on, we ignore the air resistance when studying the projectile motion.

Horizontal launch:A projectile that is being launched horizontally is shown below. First, the ball is given an initial velocity (by kicking) so that it moves horizontally on the platform. Then, it leaves the platform and becomes a projectile. The projectile follows a curved path as shown. You can put the whole path of a projectile on a plane that is perpendicular to the earth's surface. We take that plane as the $x$-$y$ plane, with the $x$ axis in the horizontal direction and the $y$ axis in the vertical direction.

The velocity of the projectile at different times are shown as a straight line with an arrow in red. The arrow shows the direction of the velocity at a given time. The initial velocity of the projectile is the velocity of the ball when it becomes a projectile. The direction of the initial velocity is horizontal, in the $+x$ direction as the ball leaves the platform when it moves horizontally. The speed (magnitude of velocity) of the projectile increases continuously until it hits the ground, shown as increasing length of the red line. As shown in the figure, the direction of the velocity changes continuously.

Projectile launched at an angle from the horizontal:A projectile that is launched at an angle from the horizontal is shown below.

Here, the projectile is not launched horizontally but at an angle from the horizontal. Once the projectile is launched, its speed decreases until it reaches the highest point. At the highest point, the speed is minimum (not zero, reason you will see later). And after that point, the speed increases until the projectile hits the ground. The direction of velocity of the projectile also changes continuously. At the highest point, the direction of velocity is horizontal as just before that point the velocity point away from the horizontal and just after that it points toward the horizontal.

The shape of a projectile's path is a parabola. You can draw the path of any projectile with a quadratic function of $y$ vs $x$.

A projectile motion is two dimensional because both the horizontal position ($x$) and the vertical position ($y$) of the projectile changes with time. So, the vectors - displacement, velocity and acceleration, have an $x$ component, and a $y$ component. The $x$ component is the horizontal component and the $y$ component is the vertical component.

We take, $t=0$ as the initial time of the projectile, which is the starting time of the projectile. So, for example, $t=2 s$ refers to the time after $2s$ of the launch of the projectile. In the figure below, the projectile is at point P at a time $t$. $\Delta x$ and $\Delta y$ are the horizontal and the vertical displacement of the projectile at P. $v_x$ and $v_y$ are the horizontal and the vertical velocity at P, and $a_x$ and $a_y$, are the horizontal and the vertical acceleration. P can be any point on the path of the projectile, if P is the end point, then $t$ represents the total time of flight.

Galileo studied the projectile motion in the 16th century. He showed that a projectile motion can be described by considering the horizontal and the vertical motion separately. This was well before the advent of vectors as the vectors were introduced only in the 19th century. We use vectors (displacement, velocity and acceleration) to describe the projectile motion. Since the $x$ and the $y$ components of a vector is independent of each other, we can study the horizontal motion with the horizontal components: $\Delta x$, $v_x$ and $a_x$. And the vertical motion with the vertical components: $\Delta y$, $v_y$ and $a_y$. So, we consider the 2-D projectile motion as two independent 1-D motion: one along the $x$ axis (horizontal motion) and the other along the $y$ axis (vertical motion). And for each 1-D motion, we will use the 1-D kinematic equations.

Acceleration of a projectile: Gravity is the only force acting on a projectile (ignoring the air resistance). Gravity pulls the object vertically downwards. So it affects only the vertical motion, and it does not have any effect on the horizontal motion. That means, the gravity cannot change the velocity of a projectile in the horizontal ($x$) direction. So, the horizontal velocity, $v_x$ is constant.

$v_x=constant.$

Since the horizontal velocity is constant, the horizontal acceleration is zero:

$a_x=0$

Gravity changes the vertical velocity, so, there is a vertical acceleration, which is the acceleration due to gravity. If we take the positive upward, then

$a_y= - g$

Initial velocity: All projectiles have a launching velocity, called initial velocity. We take the initial velocity as $\vec v_0$. Since we work with the horizontal and the vertical motion separately, we need the components of $\vec v_0$. We take $v_{0x}$ and $v_{0y}$ are the $x$ and the $y$ components of the initial velocity.

If a projectile is launched horizontally, then $\vec v_0$ is in the $+x$ direction. Since the vector is on the $x$ axis, its $y$ component is zero, and the $x$ component is $+v_0$ because the vector is in the $+x$ direction, where $v_0$ is the magnitude of the initial velocity.

$v_{0x}=v_0$  and   $v_{0y}=0$.

If a projectile is launched at an angle $\theta$ from the horizontal direction, then A projectile launched at an angle $\theta$ from the horizontal. $v_{0x}=v_0 \cos\theta$  and
$v_{0y}=v_0 \sin\theta$

Note that if the angle $\theta$ is from the vertical direction (from the $y$ axis), then you need to switch the sine and cosine in the equations.

#### Horizontal motion of a projectile

Since the horizontal velocity of a projectile is constant, the velocity at any time of a projectile is same as the initial velocity:

$v_x=v_{0x}$

where $v_x$ is the horizontal velocity at (any) time $t$.

Since horizontal velocity is constant, horizontal acceleration of a projectile is zero:

$a_x=0$

Now we need an equation for the horizontal displacement, $\Delta x$.

Take the second or the third kinematic equation and put a subscript $x$ for the velocity and the acceleration to distinguish them from the vertical components. And substituting, $a_x=0$ in the equation(s) you will get,

$\Delta x=v_{0x}t$.

In a 1-D free fall motion, you saw that the velocity of the object is zero at the highest point. But in a projectile motion, the velocity of the projectile is not zero at the highest point, because the $x$ component of velocity is constant and is always there.

#### Vertical motion of a projectile

The vertical motion of a projectile is exactly the same as the 1-D free fall motion because the vertical acceleration of the projectile is just the acceleration due to gravity. If we take the positive upward, then the vertical acceleration is

$a_y=-g$

We will use the same equations that we used for the 1-D free fall motion for the vertical motion of a projectile with a small modification, we will add a subscript $y$ to the velocity to distinguish it from the horizontal velocity. Also you can always keep the positive upward. So, we have the following equations for the vertical motion,

$v_y=v_{0y}-gt$

$\Delta y=v_{0y}t-\frac{1}{2}gt^2$

$v_y^2=v_{0y}^2-2g\Delta y$

Since the vertical motion is the same as the free fall motion, at the highest point, the vertical velocity of the projectile is zero. i.e.,

At the highest point, $v_y=0$.

#### Overall velocity at time $t$

We have, $v_x$ and $v_y$, those are the $x$ and the $y$ component of velocity at time $t$. Now, the question is what will be the velocity (i.e., the overall velocity, not the components) of the projectile at a time, $t$? Since velocity is a vector, we have the following equations for its magnitude and direction.

Magnitude of the velocity (speed of the projectile) at time $t$:

$v=\sqrt{v_x^2+v_y^2}$.

And the direction of velocity, $\theta$ at time $t$ is given by

$\tan \theta =\dfrac{v_y}{v_x}$.

Speed at the highest point: At the highest point, $v_y=0$,but since $v_x$ is always there as it is constant, the speed of the projectile at the highest point is

$v=\sqrt{v_x^2+0^2}$

$=v_x=v_{0x}$

### Relative velocity

The velocity of an object depends upon the choice of the frame of reference. When you say, the velocity of your car, you mean the velocity relative to the ground, i.e., the frame of reference is fixed at the ground. If you are traveling in a train, for you, the train is not moving, but whatever on the outside of the train appears to move. But, for a person on the ground, you are moving with the train. So, velocity is relative as we always measure the velocity with respect to some reference, and the velocity being measured is different for different reference frame.

Assume, you are at point A, on the bank of a river with water flowing from west to east. Now, you want to go to the other side of the river, to a point B that is exactly opposite to what you are now.

If you start your boat perpendicular to the river, you cannot reach the point B, as the velocity of the water stream is added to the velocity of the boat, and you will end up downstream at some point C, and your actual travel path is shown as the yellow line.

There are two velocities for the boat here, one is the velocity reference to the river bank (shore) and the other is the velocity with respect to the water. If you are in the boat, for you, the boat (and you) is moving perpendicular to the river. Since for you, water is still as you are moving with the water, like a person in the train. So, the velocity that you see is the velocity of the boat with respect to the water or the velocity in still water. But a person on the shore, see a different velocity for the boat. For him, the boat is not moving perpendicular to the river but at an angle from the north, on the yellow path. So, different reference frames see different velocities.

So, in order to reach the point B, you need to start at a certain upstream angle, i.e., at an angle from the north toward the upstream (west) and with a certain speed. For a correct angle and speed, the boat will have a velocity component exactly opposite to the velocity of the water stream that cancels the water velocity on the boat, so that the boat can travel exactly in the northerly direction to reach the point B.

Let us take $v_w$ is the velocity of the water, $v_{bw}$ is the velocity of the boat in still water and $v_{bs}$ is the velocity of the boat with respect to the shore (bank). By putting together all the velocity vectors, you can form a right angled triangle.

If the boat heads directly across, it will end up at C, and you will have

If the boat heads at an upstream angle $\theta$, it will end up at B, and you will have

When solving problems, first step is to draw one of the triangle depends on the question. So, you will have a right angled triangle. From the right angled triangle, you can solve the unknown velocity(s) or the angle, from the known quantities using the sine, cosine or the tangent or by using the Pythagorean theorem.

Some time, you may want to find the travel time, for that you need to know how wide the river is.

If $d$ is the width of the river, then the time to reach the other side is

$t=\dfrac{d}{v}$ ; [from the definition of velocity, v = d/t]

where $v$ is the velocity of the boat that is perpendicular to the river : if the boat heads north directly, then $v=v_{bw}$; but if it heads at an upstream angle, then $v=v_{bs}$. We use the perpendicular velocity because the width is perpendicular to the river.

If the boat heads directly across, and you want to find the downstream distance, BC, i.e., $x$, you will use the equation,

$x=v_w t$ ; we used $v_w$ because that is the velocity along the $x$ direction.