What is Projectile Motion?

A ball thrown at an angle to the horizontal and then allowed to fall freely under the effect of gravity alone is an example of Projectile Motion. Galileo was the first person ever to describe projectile motion, accurately. He segregated motion into horizontal and vertical components.

If the projection angle is other than 90^o, the ball follows a curved paths shown here in Fig 1. Note that if the projection angle is 90^o with horizontal, the ball follows a vertical path, as shown in Fig 2.

Yet another projectile is horizontal, if the angle of projection is zero. Refer Fig 3.

How is the angle of the projectile measured?

A projectile is thrown with an initial velocity, u from the point of projection. The angle, θ that the initial velocity, u makes with the horizontal is its projection angle .

In the given representative diagram of projectile motion of a cannon ball fired from the cannon, the initial velocity is u making an angle of projection θ with the horizontal.

Mechanics of Projectile Motion

A projectile follows a two-dimensional motion. If we take all the components of all vectors involved in the motion, vertically or horizontally, the motion can be studied as a combination of two simultaneous one-dimensional motions.

An important point to note here is that the vertical and horizontal components are independent of each other. An example for this is: Try dropping a stone and throwing another one horizontally forward at the same time from the edge of the cliff. They both hit the ground together! The forward motion does not affect the downward fall of the stone.

Components of Projectile Motion

In reference to Fig 4, where an oblique projectile is thrown, let’s consider the horizontal and vertical motions separately. Along the horizontal or the x-axis, there is no slowdown in the projectile motion in the absence or negligence of air resistance. The velocity along the x-axis remains the same as the x-component of the initial velocity, u_.  Let this be u_x . The only acceleration that affects the projectile motion is g acting vertically downwards due to gravity. Thereby the x-component of this acceleration is a_x = 0. So, it implies that the horizontal part of projectile motion is a uniform motion with a constant velocity of u_x .

In the vertical or y-axis, the projectile experiences an acceleration a_y = g along the y-direction. So, it implies that the vertical motion of the projectile is accelerated. In Fig 4, the body moves along O to A, Its state of deceleration, till its vertical component of velocity, u_y is reduced to zero. This is the highest point of the flight at A. Thereafter, along A to B, the body gets accelerated where its vertical velocity v_y keeps increasing.

The various kinematic equations describing the projectile motion are the following:

Equations of Projectile Motion

Resolving the initial velocity along horizontal and vertical axes:

u_x = ucosθ

u_y = usinθ

For the projectile’s position at any point P during the flight, the displacement coordinates are x and y along horizontal and vertical axes, respectively.

Along x-axis of the projectile equation

x = utcosθ (uniform motion)

Along y-axis projectile equations:

y = utsinθ - gt²/2 (uniformly accelerated motion)

Substituting for t,

The equation of motion for the projectile is:

$y=x\tan \theta - \frac{g}{2u^2{\cos }^2\theta }{x}^2$

Clearly, this is a quadratic equation in x, implying that the path followed by the projectile is parabolic.

Time of flight is the time the projectile takes to strike the ground back at the same height as that of the point of projection. Note that at point B (Fig 6), the vertical displacement is zero.

Time of flight,  $T = \frac{2u\sin \theta }{g}$

Range of the projectile is the total distance, OB (Fig 6) traveled in the horizontal direction before striking the ground (that’s at the same vertical level as that of the initial point of projection)

Range of the projectile, $R = \frac{u^2\sin 2\theta }{g}$

Height of the projectile h (Fig 6) is the maximum vertical distance attained by the projectile before it begins its descent.

Height of the projectile, $H = \frac{u^2\sin 2\theta }{2g}$

Range of projectiles plays a significant role primarily for military purposes—such as aiming cannons on the enemy locations. Consequently, analyzing the range of projectiles can help understand interesting phenomena, such as the orbits of satellites around the Earth.

From the formula, $R = \frac{u^2\sin 2\theta }{g}$, we see that the range of a projectile depends upon its initial velocity and the angle at which the projectile is launched at the onset. A canon can be launched mostly at a fixed speed, wherein the maximum range is obtained with θ⁰ = 45º. However, this is true, provided air resistance is ignored. In the presence of air resistance, the maximum angle is approximately 38º in order to produce maximum range for a given initial velocity. An interesting point to know here is that for every projection angle, except 45º, two angles give the same horizontal range! These two angles are complementary to each other.

As per the formula, range also depends upon acceleration due to gravity at a place. Alan shepherd could strike the golf ball through an extensive range on moon! This was possible because the acceleration due to gravity on the moon is about one-sixth that on Earth.

The next question arises, what if the projectile is launched on Earth’s surface from a height with a velocity and angle such that its range is very large and is comparable to the circumference of the Earth?

This is how a satellite is launched in Earth’s orbit. A satellite is a projectile launched from a great height from the Earth with a great speed, such that it has a very large range. As the satellite falls towards the Earth under the effect of gravity, its range being large, the Earth curves away below the projectile. In effect, the projectile is always falling towards the Earth, but unable to reach back the ground as the Earth’s surface curves down. As a result, the satellite begins to orbit around the Earth!

Context and Applications

Cricket, basketball, rifle shooting, gymnastics or even shotput and javelin throws are few of the sports where the principle of projectile motion needs to be well understood and practiced for a sportsperson to excel in his or her game.

Understanding of basics of Projectile motion assists in forensic based crime investigation as well! The nature of the ballistic projectile used at the crime scene decides the speed with which the bullet or cannon is fired. The basics of projectile motions help identify the probable location of the strike that can help in the crime scene investigations.

Fundamentally, free movement of an object without any support is example of projectile motion.

This topic is significant in the professional exams for both undergraduate and graduate courses, especially for 

• Bachelors in Science (Physics) 
• Masters in Science (Physics)

Practice Problems

Q1. A person in a car moving forward to the right fires a gunshot vertically up. In the absence of the air friction, where will the bullet land?

1. Behind the car
2. Ahead of the car
3. In the barrel of the gun

Answer: In the barrel of the gun

Q2. If in case of projectile motion, horizontal range is n times the maximum height H, then the angle of projection is

1. tan^-1(n/4)
2. tan^-1(4/n)
3. tan^-1(4n)
4. 4/n

Answer: tan^-1(4/n)

Q3. A projectile fired from the ground follows parabolic path and its speed at the maximum height position is zero.

1. True
2. False

Answer: False

Q4. A man can throw a ball through a maximum range of 16 m. The maximum height to which the ball will rise is:

1. 16 m
2. 10 m
3. 8 m
4. 4 m

Answer: 4m

Q5. Weight of a body during the projectile motion is zero.

1. True
2. False

Answer: False