# Projectile motion

##### The acceleration due to gravity near the earth’s surface

A projectile is any object that has been given an initial velocity and left to move under the action of gravity alone.

Specifically, we study the motion of objects near the earth’s surface over distances over which the acceleration due to gravity does not change by much. At a later stage we learn to include the effects of air friction.

The force of gravity that the earth exerts on a small object of mass m is given by Newton’s law of gravitation:

(1)

where G ( 6.67 x 10^{-11} N m^{2} / kg^{2} ) is the universal gravitational constant, M is the mass of the earth and r is the distance of the object from the center of the earth. Let the radius of the earth be R and the object’s height above the earth’s surface be h. Then r = R + h. Considering that the radius of the earth is some 6700 km, h for ordinary projectiles like stones, cannonballs, basketballs, missiles is negligible compared to R. So the force of gravity on any of these objects is practically independent of their height above the earth’s surface.

(2)

where a is the acceleration of the object. m cancels on both sides, giving us a number that is independent of the accelerating object.

(3)

This number evaluates to approximately 9.8 m/s^{2}. This acceleration due to gravity of any object, be it a feather or a cannonball, is denoted g. It varies slightly over the surface of the earth.

A particle that is projected with an initial velocity at some angle will travel in a parabolic arc .

Figure 1

##### How do we predict the motion of an object if we know its initial velocity?

The most important thing to understand here is that we can treat the motion in the x and the y directions independently. That is, *what happens in the x direction does not in any way affect what happens in the y direction*.

We separate the projectile motion into two, the motion of the particle’s x-coordinate and its y-coordinate.

Here is an example problem:

A particle is projected from the ground with the initial velocity of 40 m/s at an angle of 60^{0} to the horizontal. How high does it reach? How far away and with what velocity does it hit the ground again?

The initial velocity of 40 m/s at an angle of 60^{0} is first split into its x- and y- components.

(4)

(5)

Figure 2

The accelerations in the two directions are: a_{x} = 0, a_{y} = -g = -9.8 m/s^{2} ( we have chosen the upward direction to be the positive y- direction.)

To make our calculations easy, from now on we will take g to be 10 m/s^{2}.

We now proceed to apply the three equations we learned for constant acceleration in 1-D, separately to each of the two directions:

(6)

(7)

For our problem,

for the x-direction we have,

(8)

That is, the x-component of the velocity stays equal to its initial value. This is because there is no force in the x-direction.

And for the y-direction we have,

(9)

At the top most point the y-component of the velocity is zero. Just before it reaches the top, its y-coordinate is increasing, making v_{y} positive and just after it reaches the top its y-coordinate is decreasing, making v_{y} negative. At the very top, v_{y} is momentarily zero.

(10)

Now that we know the time it takes to reach the top, we can calculate the maximum height : this is the displacement in the y-direction at t = 2√3 s.

(11)

To find where it lands, we calculate its x-displacement at the instant the y-displacement is zero. When it comes back to the same level, the change in the y-coordinate is zero.

(12)

The two solutions to the quadratic equation tell us that the y-displacement is zero at t = 0 and at t = 4√3 s. t = 0 is the initial instant -- the instant at which the projectile was fired. Obviously, the y-displacement is zero at that instant. So, t = 4√3 is the solution we are interested in and gives us the time of travel.

This shows that the time taken to go up is equal to the time taken to come down: 4√3 = 2 * 2√3.

**Q**. Is that always true for all projectiles?

**A**. Yes, assuming that air friction is absent.

During the 4√3 seconds, the x-displacement is 20 x 4√3 = 80√3 m. This is called the range.

The final velocity is obtained from,

(13)

Comments:

1. The x and the y directions are treated independently, as though we are solving two 1-D problems. This is true of all vector equations that we deal with. *Each vector equation is equivalent to 3 ordinary( scalar ) equations ( in 3-D), one for each direction, x, y and z* . The projectile’s trajectory, however, lies in one plane (why?), and so we need only the x and the y components and no z component.

**F** = m**a** is a vector equation. We solve it by setting up a coordinate system, splitting the vectors into their components along each direction, and then solving the equation for each direction separately.

Similarly, **v** = **u** + **a**t is also a vector equation. We solved each of its component equations separately.

2. I remind you again: the equations that we use are valid only for the case of constant acceleration. This is true as long as the projectile does not go too high (the acceleration due to gravity decreases with height) or so far that the earth cannot be treated as flat.

We have also neglected wind resistance.