Projectile Motion

Projectile and projectile motion

When an object is thrown with a certain initial velocity and that moves under the effect of gravity alone, we say that object is a projectile and the motion is projectile motion. The path of that motion is called a trajectory.

An object thrown from a height, a bullet fired from a gun, etc. are some examples of a projectile.

In real life, air resistance also affects motion but we neglect air resistance for simplicity.

Projectile thrown from the ground

Consider a projectile is thrown from the ground with some initial velocity u m/s with the angle \theta from the horizontal(ground).

projectile thrown from ground

To find all the expressions related to projectile motion, we have to use the concept of the resolution of a vector.

We resolve the velocity of a projectile at all points of time to two components: horizontal component of velocity and vertical component of velocity.

And we can write: u_x = u \cos\theta and u_y = u \sin\theta

In the absence of air resistance, the horizontal component of velocity remains constant throughout the trajectory but the vertical component of velocity is affected by gravity. So, the vertical velocity first decreases at a rate of 9.8m/s^2 during upward motion, becomes zero at the top, and finally increases at the same rate of 9.8 m/s^2 and reaches the ground.

Some notations:

  • u = initial velocity
    • u_x = horizontal component of initial velocity
    • u_y = vertical component of initial velocity
  • v = final velocity after t time
    • v_x = horizontal component of final velocity
    • v_y = vertical component of final velcoty
  • t = time taken to reach final velocity
  • g = acceeration due to gravity (g=9.8 m/s^2)
  • \theta = angle of projection

Let’s first find out expressions of distance and velocity for both horizontal and vertical components:


horizontal velocity remains constant whereas vertical velocity is affected by gravity.

Horizontal component:

a) Distance:


(1)   \begin{align*}x &= u_x t \\     x &= u\cos\theta t\end{align*}

b) Velocity

final velocity:

(2)   \begin{align*}v_x &= u_x \\  v_x &= u\cos\theta\end{align*}

Vertical Component

a) Distance


(3)   \begin{align*}y &= u_y t - \frac{1}{2}gt^2\\   y &= u\sin\theta t - \frac{1}{2} gt^2\end{align*}

b) Velocity

final velocity:

(4)   \begin{align*}v_y &= u_y - gt\\    v_y &= u\cos\theta -gt\end{align*}

from euqation 1 we can write:

(5)   \begin{align*}t = \frac{x}{u\cos\theta}\end{align*}

Substituting this t value in equation 3

(6)   \begin{align*}y &= u \sin\theta \frac{x}{u \cos\theta} - \frac{1}{2}g\left(\frac{x}{u\cos\theta} \right)^2 \\y &= \tan\theta \hspace{2mm} x - \left( \frac{g}{2u^2\cos^2\theta} \right) x^2 \\y &= Ax - Bx^2  \end{align*}

Where A = \tan\theta and B = \frac{g}{2u^2 \cos^2\theta} are constant terms.

If we plot 6, we get a parabola graph. Hence, we can say that the trajectory of a projectile is parabolic in nature.

Time to reach maximum height

We take the vertical component for this calculation, as the movement of the projectile should be upward to reach maximum height.

Note that the vertical velocity becomes zero at the maximum height, as the projectile velocity decreases at the rate of 9.8 m/s at every second and becomes zero.

from equation 4:

(7)   \begin{align*}v_y &= u \cos\theta - gt \\0 &= u \cos\theta - gt \\t &= \frac{u \cos\theta}{g}\end{align*}

    \[ \boxed{t = \frac{u \cos\theta}{g}} \]

Time of flight

Time of flight is the total time that projectile spent in the air. It is denoted by ‘T’. Remember the time to reach maximum height is equal to the time to reach the ground from that maximum height.

So, we can get the Time of flight by doubling the time to reach maximum heght.

(8)   \begin{equation*}\text{Time of flight} =2 \times \text{time to reach the maximum height}\end{equation*}

    \[ \boxed{T = \frac{2u \sin\theta}{g}} \]

Maximum Height

It is the maximum vertical distance taveled by the projectile thourghout the motion. We represent maximum height by h_{max}. Note that, the vertical component of velocity becomes zero at the maximum height.

maximum height of projectile

(9)   \begin{align*}{v_y} ^2 &= {u_y}^2 - 2gh \\0 &= \left(u \sin\theta \right)^2 - 2 g h_{max} \\0 & = u^2\sin^2\theta - 2 g h_{max} \\h_{max} &= \frac{u^2 \sin^2\theta}{2g}\end{align*}

    \[ \boxed{h_{max} = \frac{u^2 \sin^2\theta}{2g}} \]

Horizontal Range

It is the horizontal distance traveled by the projectile during its motion. We consider the horizontal component of velocity to calculate the horizontal range. It is denoted by ‘R’.

horizontal range

from equation 1:

(10)   \begin{align*}x &= u \cos\theta \hspace{2mm}t \\R &= u \cos\theta \hspace{2mm}T \\R &= u \cos\theta \hspace{2mm}\frac{2u\sin\theta}{g} \\R &= \frac{u^22\sin\theta\cos\theta}{g} \\R &= \frac{u^2\sin2\theta}{g}\end{align*}

    \[ \boxed{ R = \frac{u^2 \sin2\theta}{g}} \]

Two angles of projection for same horizontal range

We have, horizontal range:

    \[  R = \frac{u^2 \sin2\theta}{g} \]

where, \theta is the angle of projection.

If (90^0 - \theta) is another angle of projection, the range becomes,

    \begin{gather*}R = \frac{u^2 \sin2(90^0-\theta)}{g} \\R = \frac{u^2 \sin(180^0-2\theta)}{g} \\R = \frac{u^2 \sin2\theta}{g}\end{gather*}

So the horizontal velocity R for the given velocity u is the same for both angles of projection \theta and (90^0 - \theta)

Velocity of projectile at any instant

Suppose v_x and v_y are the velocity components at any instant of time t then, we can calculate the velocity at that instant of time using formula:

    \[v = \sqrt{ {v_x}^2 + {v_y}^2 }\]

where, v_x and v_y can be calculated from equations 3 and 4.

Some FAQ’s

Sharing is Caring