Projectile Motion

Motion in a plane is also known as two-dimensional motion. For instance, circular motion, projectile motion, and so on. The reference point for such motion analysis will be an origin and the two coordinate axes X and Y. In this section we will discuss the Projectile motion. 

The motion of an object thrown or projected into the air, subject only to the acceleration of gravity, is referred to as projectile motion. The object is known as a projectile, and its path is known as its trajectory. Falling object motion, as described in Problem-Solving Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion with no horizontal movement. In this section, we look at two-dimensional projectile motion, such as that of a football or other object with negligible air resistance.

The most important thing to remember here is that motions along perpendicular axes are independent of one another and can thus be analyzed separately. This was discussed in Kinematics in Two Dimensions: An Introduction, where it was demonstrated that vertical and horizontal motions are independent. The key to understanding two-dimensional projectile motion is to divide it into two motions, one horizontal and one vertical. (This is the most logical choice of axes because gravity’s acceleration is vertical, so there will be no acceleration along the horizontal axis when air resistance is negligible.)

Of course, in order to describe motion, we must consider velocity, acceleration, and displacement. We must also locate their components along the x– and y–axes. We will assume that all forces other than gravity (such as air resistance and friction) are negligible. The components of acceleration are then very straightforward:

• a_y = –9.80 m/s². (Please keep in mind that this definition assumes that the upwards direction is defined as the positive direction, if the coordinate system is arranged in such a way that the downward direction is positive, the acceleration due to gravity is positive.) 
• a_x = 0 because gravity is vertical. 

Because the accelerations are constant, the kinematic equations can be applied.

Projectile Motion

A projectile is any object thrown into space with only gravity acting on it. The primary force acting on a projectile is gravity. This is not to say that other forces do not act on it; rather, their impact is minimal when compared to gravity. A trajectory is a path taken by a projectile.

When a particle is thrown obliquely near the earth’s surface, it follows a curved path with constant acceleration toward the earth’s center (we assume that the particle remains close to the surface of the earth). The path of such a particle is known as a projectile path, and its motion is known as projectile motion.

Projectile motion is one of the most common types of motion in a plane. The only acceleration acting in a projectile motion is the vertical acceleration caused by gravity (g). As a result, equations of motion can be used separately in the X- and Y-axes to determine the unknown parameters.

• The motion of a projectile in two dimensions is divided into two parts:

1. Horizontal motion in the x-direction with no acceleration and
2. Vertical motion in the y-direction with constant acceleration due to gravity.

• The equation for projectile motion is y = ax + bx².
• To simplify calculations, projectile motion is typically calculated without accounting for air resistance.

Before understanding the derivation of the relation for projectile motion lets first introduce some terms use in it, that are: 

• Angle of Projection: The angle at which the body is projected with respect to the horizontal is referred to as the angle of projection.
• Velocity of Projection: The velocity with which the body is thrown is referred to as the velocity of projection.
• Point of Projection: A point of projection is the point from which the body is projected in the air.
• Projectile Trajectory: The path taken by a projectile in the air is referred to as the projectile’s trajectory.
• Horizontal Range: The horizontal distance traveled by the body performing projectile motion is referred to as the range of the projectile.

Consider the following example of a ball which is projected at an angle θ from the point O with respect to the horizontal x-axis with an initial velocity u:

Here,

• The point O is known as the point of projection.
• θ is the angle of  projection and
• OB = Horizontal Range.
• H is the height of the particle.
• The total time taken by the particle to travel from O to B is referred to as the time of flight.

We can use differential equations of motion to find various parameters related to projectile motion.

We know that the linear equation of motion are:

                         v = u + at

                       S = ut + 1/2(at²)

                       v² = u² + 2aS

 Applying the above equation for projectile motion the equation will be:

                       v = u – gt

                       S = ut – 1/2(gt²)

                       v² = u² – 2gS

   Where,

          u =  initial velocity

          v = Final velocity

          g = Acceleration due to gravity (Taking it -ve because gravity always work downward)

          S = Displacement

          t = Time                                                           

Total Time of Flight:

In Y direction total displacement (S_y) = 0.

so taking motion in Y direction,  S_y  = u_yt – 1/2(gt²)   [Here, u_y = u sinθ and S_y = 0]

                                            i.e.      0 = usinθ – 1/2(gt²)

                                                      t = 2usinθ/g

Total Time of Flight(t) = 2usinθ/g

Case 1:  if θ = 90°  

As we can see from the formula of Time of flight, time taken by the projectile is directly proportional to the angle of projection. For any given initial velocity(u) will be constant and g is always constant i.e. g=-9.8 m/s2.

When projectile is projected at an angle of  90° time of flight will be maximum.

So,    t_max = 2usinθ/g = 2u/g     [ sin 90° = 1]

Case 2:  if θ = 30°  

When the projectile is projected at an angle of 30° time of flight is half of the t_max .

i.e. t = 2usin30°/g = t_max/2.    [  sin30° = 1/2 ]  

Horizontal Range:

Horizontal range is a distance(OB) it can be given as:

                                          OB = Horizontal component of velocity(u_x) * Total time(t)       [ Here, u_x = u cosθ and t = 2usinθ/g]

                                   i.e.   Range(R) = ucosθ * 2usinθ/g              

As a result, the Horizontal Range of the projectile is given by (R):

Horizontal Range(R) = u²sin2θ/g                    [ Here, sin2θ = 2cosθsinθ]

Case 1:  if θ = 90°  

When projectile is projected at an angle of 90° Horizontal range will be zero, because projectile will strike at the same point where the projectile is projected. 

i.e.  R  = u²sin2θ/g = 0.    [ sin 2θ = 0, at θ = 90 ]

Case 2: if θ = 45°

When projectile is projected at 45° Horizontal Range of the projectile is maximum.

i.e. R  = u²sin2θ/g = u²/g.     [ Here, sin90 = 1 ]

Maximum Height:

It is the highest point of the particle (point A). When the ball reaches at the point A, the vertical component of the velocity (V_y) will be zero. 

                                          i.e.    0 = (usinθ)²  – 2gH_max                            [ Here, S = H_max , v_y = 0 and u_y = u sin θ ]

Therefore, the Maximum Height of the projectile is given by (H_max):

Maximum Height (H_max) = u²sin²θ/2g

Case 1: if θ = 90°

If we project a projectile at an angle of 90° it achieves maximum height (H_max).

i.e.    H_max  =  u²sin²θ/2g = u²/2g.        [Here, sin²90 = 1 ]

Case 2: if θ = 45°

When projectile is projected at an angle of if 45°, height of projectile is half of its maximum height (Hmax).

i.e.    H  = u²sin²θ/2g = (1/2)u²/2g = Hmax/2.           [ sin²45° = 1/2 ]

We can also say that if the projectile angle is 45° than Horizontal range of projectile will be 4 time the height of projectile.

i.e.   H = u²/4g = R/4       [ Here, Horizontal range at θ = 45°, R = u²/g ]

or    R  = 4H.

The equation of Trajectory:

Equation of the trajectory is a path followed by the particle during the projectile motion. The equation is:

y = x tanθ – gx²/2u²cos²θ

This is the equation of projectile motion it is similar to the parabola( y = ax + bx²)so we can say that projectile motion is always parabolic in nature.

Examples of two-dimensional motion are as follows,

• Throwing a ball or launching a cannonball
• A billiard ball’s movement on a billiard table.
• The motion of a gun shell being fired.
• The earth’s rotation around the sun.
• When a ball is thrown from a moving car the path traverse by the ball is projectile motion.
• A fielder throws the ball towards the wickets in a cricket match.
• A bullet is fired at a long-distance target.

Some Applications of the Projectile motion:

• A rocket or missile is a more complex type of projectile application in modern life.
• Projectiles are commonly used by athletes, particularly in the javelin throw, shot put, discus, and hammer throw, among others.
• Archery and shooting also make use of projectiles.

Sample Problems

Problem 1: What is projectile? Prove that the path of a projectile is parabolic.

Solution:

Projectile: A projectile is any object thrown into space with only gravity acting on it.

We know that the equation of projectile is, y = x tan θ – gx²/2u²cos²θ comparing the equation with y = ax + bx² 

Here, a = tanθ, and b = – g/2u²cos²θ. 

The above equation of trajectory is similar to a parabola. So it is proved that the path of a projectile is parabolic.

Problem 2: At what angle projectile should be projected so that the height and Range of the projectile will be equal?

Solution:

If height and horizontal range will be equal than, H = R.

i.e.  

u²sin²θ/2g = u²sin2θ/g   

or

sin²θ = sin2θ                                                                     [Here, sin2θ = 2 sinθ cosθ and tanθ = sinθ/cosθ]

tanθ = 4                   

So,            

θ = tan^-1(4)

Problem 3: Define horizontal range and find the range of a projectile thrown at 98 m/s with an angle of 30 degrees from horizontal? (g =9.8 m/s²)

Solution: 

Horizontal Range- The horizontal distance traveled by the body performing projectile motion is referred to as the range of the projectile.

Horizontal Range, R = u²sin2θ/g 

                                = (98)² × (60°) / 9.8 

                                = 490√3 m

Problem 4: Name the physical quantities which will remain unchanged during the projectile motion?

Solution: 

Velocity, vertical component of velocity momentum, kinetic energy, and potential energy remain unchanged during the projectile motion.

Problem 5: What is the maximum height attained by a ball of mass 100 g projected at an angle of 30° from the ground with an initial velocity of 11 m/s and an acceleration due to gravity of g = 10 m/s²?

Solution:

We know that the formula for maximum height is,

H = (usinθ)²/2g.

Given,  

u = 11 m/s, θ = 30°, g = 10 m/s²

Hence,  

Putting the values we get,

H = 1.5125 m 

Problem 6:A football is launched at a 45° angle from the ground with an initial velocity of 10 m/s; the gravity acceleration is g = 10 m/s². What is the time of flight?

Solution:

We know that the formula for calculating the time of flight is, 

t = 2(usinθ/g)

Given, θ = 45°, u = 10m/s, g = 10m/s²

Putting the values we get,

t = 1.4142 s

Problem 7: A projectile is projected from point O at an angle of 30° with an initial velocity of 30 m/s. The projectile hits the ground at point M. (Consider acceleration of gravity g = 10m/s²) Find the following:

1. What is the total time of flight?
2. What is the Horizontal Range of the projectile (OM)?
3. What is the maximum height of the projectile?

Solution:

Given,

Initial velocity u = 30m/s.

The angle of projection, θ = 30°.

 1. Total time of flight

We know that the total time of flight by the projectile is given by-

 t = 2usinθ/g

Putting the given values,   

t = 2 × 30 sin30°/10 

  = 3 s

 2. Horizontal Range

We know the formula for horizontal range is:  

R = u²sin2θ/g.

Putting the values we get,     

R = (30)² sin60° /10

   = 45 √3 m.

3. Maximum Height

Maximum height of the projectile is given by the formula:   

H_max = u²sin²θ/2g

Putting the values we get,    

H_max = (30)²sin²30°/2 × 10

        = 11.25 m.