What is Projectile Motion?

Definition: The motion of any object after being thrown in any direction from a point on the earth's surface or near the earth's surface is called projectile motion.

Examples: The motion of a football after being kicked, the motion of a cannonball after being fired from a cannon, the motion of a basketball, the motion of a golf ball, etc.

If an object is thrown horizontally without being thrown vertically upwards, the object no longer returns to the point of projection, rather than it moves slightly away from the point of projection.

It is actually thrown this way to move the object a little farther horizontally. For example, kicking a football. The object that is thrown here is called a projectile and the path followed by the projectile is known as trajectory.

Note:

Projectile: An object that is thrown or fired is called a projectile.

Trajectory: The path followed by a projectile is known as trajectory.

Formulas for Projectile Motion

Observing the Projectile Motion we can determine:

1. Equation of path of a projectile

2. Time of flight of a projectile (Tf)

3. Maximum height of a projectile (Hm)

4. Horizontal range of a projectile (R)

What is Projectile Motion? Definition, Concepts, and Formulas

In the above projectile motion, the point O is called the point of projection, θ is the angle of projection, and OB = horizontal range. The total time taken by the object from reaching O to B is called the time of flight.

Now we can use the equations of motions to find out different parameters related to projectile motion.

${\color{Red} v=u+at\rightarrow (1)}$

In terms of components:

${\color{Blue} v_{x}=u_{x}+a_{x}t\rightarrow (2)}$

${\color{Blue} v_{y}=u_{y}+a_{y}t\rightarrow (3)}$

For the 2-dimensional motion the formula for the position vector 'r' ( Read more from what is position vector and displacement vector?)

${\color{Red} r=r_{0}+ut+\frac{1}{2}at^{2}\rightarrow (4)}$

In terms of components:

${\color{Blue} x=x_{0}+u_{x}t+\frac{1}{2}a_{x}t^{2}\rightarrow (4)}$

${\color{Blue} y=y_{0}+u_{y}t+\frac{1}{2}a_{y}t^{2}\rightarrow (6)}$

Here, u = initial velocity, v = final velocity, a = acceleration due to gravity, r = position vector, t = time.

In a Projectile Motion, there are two independent components of rectilinear motions:

Along the x-axis: Responsible for the horizontal motion of the particle.

The components of the initial velocity (ux) = ucos𝜃

the component of acceleration (ax) = o

Along the y-axis: Responsible for the vertical motion of the particle.

The components of the initial velocity (uy) = usin𝜃

The component of acceleration (ay) = -g

When a particle is projected into the air at some speed, the only force acting on it during its flight in the air is gravity (g). This acceleration only works vertically downwards.

There is no acceleration in the horizontal direction. So the velocity of the particles in the horizontal direction remains constant.

Now if we take the initial position of the object to be the origin of the reference frame, we have: x0 = 0, y0 = 0

Then from equation no-(5, 6), we get

$x=u_{x}t=ucos\theta t$

[∵ x0 = 0, ux = ucos𝜃 , and (ax) = o]

$y=usin\theta t-\frac{1}{2}gt^{2}$

[∵ y0 = 0, uy = usin𝜃 , and ay = -g]

The components of velocity at time t can be obtained using equation no-(2, 3)

$v_{x}=u_{x}=ucos\theta$

[∵ ux = ucos𝜃]

$v_{y}=usin\theta t-gt$

[∵ uy = usin𝜃 , and ay = -g]

Equation of path of a projectile :

What type of trajectory a projectile followed can be seen through expressions of x and y as given above.

$x=u_{x}t=ucos\theta t$

$or \ t=\frac{x}{ucos\theta }$

$Now \ from \ y=usin\theta t-(\frac{1}{2})gt^{2}$

$We \ get \ y=usin\theta \times \frac{x}{ucos\theta }-(\frac{1}{2})g (\frac{x}{ucos\theta })^{2}$

${\color{Red} Or\ y=xtan\theta -\frac{gx^{2}}{2u^{2}cos^{2}\theta }}$

This is the Equation of path/trajectory of a projectile motion. By comparing this equation with y = a x + b x2, (Here a and b are constants) we can say that the path of a projectile is parabolic in nature.

Time of flight of a projectile (Tf) :

How much time does a projectile take to reach the maximum height can be determined through expressions of vx and vy as given above.

$v_{y}=usin\theta -gt_{m}=0$

Here tm is the time taken by a projectile to reach the maximum height where vy = 0

$So \ t_{m}=\frac{usin\theta }{g}$

Now if we take the Tf as the total time during the flight of the projectile then the Tf will be twice the Tm (where y = 0 and x = max).

$T_{f}=2t_{m}$

${\color{Red} T_{f}=\frac{2usin\theta }{g}}$

Maximum height of a projectile (Hm) :

The equation of maximum height Hm can be determined by substituting t = tm in the expressions of y as given above.

$y=usin\theta t-\frac{1}{2}gt^{2}$

$y=H_{m}=(usin\theta )(\frac{usin\theta }{g})-\frac{g}{2}(\frac{usin\theta }{g})^{2}$

$H_{m}=\frac{u^{2}sin^{2}\theta }{g}-\frac{u^{2}sin^{2}\theta }{2g}$

${\color{Red} H_{m}=\frac{u^{2}sin^{2}\theta }{2g}}$

Horizontal range of a projectile (R) :

The horizontal range (x = R) is the distance travelled by a projectile from its initial position (x = y = 0) to the position where it passes y = 0 during its fall.

It is the distance traveled during the time of flight Tf. It can be determined by substituting t = Tf in expressions of x as given above.

$x=ucos\theta t$

$x=R=ucos\theta T_{f}$

$R=ucos\theta \times (\frac{2usin\theta }{g})$

$R=\frac{u^{2}2sin\theta cos\theta }{g}$

${\color{Red} R=\frac{u^{2}sin2\theta }{g}}$

This equation shows that for a given projectile motion where velocity is u, R will be maximum when sin2θ is maximum, i.e., when θ = 45. The maximum horizontal range will be

${\color{Red} R=\frac{u^{2}}{g}}$

Also, read about uniform circular motion...

Follow Us:

What is Projectile Motion? Definition, Concepts, and Formulas

What is Projectile Motion?

Formulas for Projectile Motion

Contact Form