# Projectile Motion

When first getting a grip on classical mechanics, it is important to digest certain seminal examples. One of these examples is that of a **projectile**: an object that is subject only to the force of gravity.

Since it is only subject to gravity, the projectile cannot be in contact with any surface. Hence, projectiles are almost always in non-propelled flight. The most classic example is a ball that has been thrown into the air.

Studying *projectile motion* allows for full application of kinematics, various equations of kinematic-motion and vector geometry.

#### Contents

## Examples of projectile motion

- A glass accidentally falling off a table.
- A phone tossed into a bed.
- A missile deployed from a military aircraft from level flight.
- A Javelin thrown by an athlete.

## Assumptions of Projectile motion

- There is no friction due to air.
- The effect due to the curvature of earth is negligible.
- The effect due to the rotation of earth is negligible.
- The entire trajectory is near the surface of the Earth.
- For all points of the trajectory, the acceleration due to gravity
*g*is constant in magnitude (\(9.8 \dfrac{m}{s^2}\)) and direction (toward the Earth).

## Velocity and Acceleration in Projectile Motion

Consider the diagram shown below. It shows that the velocity of a projectile is changing. Remember that the velocity is always tangential to the path. As the path curves, velocity also changes direction.

When the projectile is in air and air resistance is neglected then the only force acting on it is gravitational force. At all times the acceleration of the particle is constant in magnitude as well as direction i.e. *g* downwards.

## Principle of Physical Independence of Motions

Projectile motion is a planar motion in which at least two position coordinates change simultaneously.

**Principles of Physical Independence of Motions**

1) The motion of a projectile is a two-dimensional motion. So, it can be discussed in two parts. Horizontal motion and vertical motion. These two motions take place independent of each other. This is called the principle of physical independence of motions.

2) The velocity of the projectile can be resolved into two mutually perpendicular components: the horizontal component and the vertical component.

3) Acceleration changes velocity. If acceleration in a particular direction is zero then velocity in that direction remains the same. Thus, in projectile motion, the horizontal component of velocity remains unchanged throughout the flight. The horizontal motion is a uniform motion.
4) The force of gravity continuously affects the vertical component, so the vertical motion is a uniformly accelerated motion.

Projectiles can be thrown in various ways: on a level ground, from a high tower to ground, from an aeroplane etc. The following section discuss some cases in detail.

## Oblique Projectiles on Level Ground

In the diagram, a particle is thrown with speed *u* at an angle \(\theta\) with the horizontal.

An analysis of the motion of the projectile starts by breaking the components of initial velocity and acceleration into horizontal (along x-axis) and vertical (along y-axis) components. Breaking the velocity and acceleration into independent components helps to study a 2-D projectile motion as two independent 1-D motions.

The horizontal component of velocity \( {u_x} = u\cos \theta \)

The vertical component of velocity \( {u_y} = u\sin \theta \)

The horizontal component of acceleration \( {a_x} = 0 \)

The vertical component of acceleration \( {a_y} = - g \) (Negative sign shows that the acceleration is in a downward direction)

**Time of Flight**

The total time spent in the air is \(T = \dfrac{{2u\sin \theta }}{g}\).

\[{S_y} = {u_y}t + \frac{1}{2}{a_y}{t^2}\] The projectile lands back at the same height, so the displacement in vertical direction is zero.

\[\begin{align} 0 &= u\sin \theta T - \dfrac{1}{2}g{T^2}\\ T &= \dfrac{{2u\sin \theta }}{g} \end{align}\]

**Important point**: The time of flight is independent of the horizontal component of velocity. The faster a projectile is thrown up, the longer it will stay in the air.

**Maximum Height**

The maximum height a projectile reaches above its release point is \({H_{\max }} = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}\).

AVOID THIS PITFALLThe velocity at the high point in projectile motion is not zero. Although the vertical component of velocity is 0.For motion in the vertical direction. \[v_y^2 = u_y^2 + 2{a_y}{S_y}\]

At the highest point, vertical component of velocity is zero.

\[\begin{align} 0 &= {(u\sin \theta )^2} - 2g{H_{\max }}\\ {H_{\max }} &= \frac{{{u^2}{{\sin }^2}\theta }}{{2g}} \end{align}\]

- The maximum height is independent of the horizontal component of velocity. The faster a projectile is thrown upwards, the higher it will go in an upward direction. I.E. the longer it will resist the downward pull of gravity.
- Both time of flight and maximum height depend on the vertical component of velocity, thus the relation between them can be expressed as

\[\dfrac{{{H_{\max }}}}{{{T^2}}} = \dfrac{g}{8}\]

**Horizontal Range**

The net displacement in the horizontal direction for an object with no vertical displacement is \(R = \frac{{{u^2}\sin 2\theta }}{g}\)

For motion in horizontal direction,

\[{S_x} = {u_x}t + \frac{1}{2}{a_x}{t^2}\]

Acceleration in horizontal direction is zero

\[\begin{align} R& = (u\cos \theta )T\\ R& = (u\cos \theta )\frac{{2u\sin \theta }}{g}\\ R& = \frac{{2{u^2}\sin \theta \cos \theta }}{g} = \frac{{{u^2}\sin 2\theta }}{g} \end{align}\]

- The horizontal range depends upon both the horizontal and vertical component of velocity.
- For a specified speed of projection, the range will max out at an angle of projection equal to \(45^\circ\).
- For projectiles moving at equal speed, the range will be equal when both projectiles have complementary angles of projection.
- The relation between range, maximum height and time of flight is \(R\tan \theta = \frac{1}{2}g{T^2} = 4H\).

**Equation of Trajectory**

The equation of the path followed by a projectile is \(y = x\tan \theta \left( {1 - \frac{{gx}}{{{u^2}\sin 2\theta }}} \right)\)

The equation of a curve in two dimensional space is a relation between x and y coordinates. By using equations of motion in horizontal and vertical directions respectively we can find the relations between x,t and y,t. Then by eliminating t we can find the relationship between x and y.

For motion in x direction \[\begin{array}{l} {S_x} = {u_x}t + \frac{1}{2}{a_x}{t^2}\\ x = u\cos \theta \,t \end{array}\]

\[t = \frac{x}{{u\cos \theta }}\]For motion in y direction \[\begin{array}{l} {S_y} = {u_y}t + \frac{1}{2}{a_y}{t^2}\\ y = u\sin \theta \,t - \frac{1}{2}g{t^2} \end{array}\]

Eliminating

tfrom the equations, \[\begin{array}{l} y = u\sin \theta \frac{x}{{u\cos \theta }} - \frac{1}{2}g{\left( {\frac{x}{{u\cos \theta }}} \right)^2}\\ y = x\tan \theta \left( {1 - \frac{{gx}}{{{u^2}\sin 2\theta }}} \right) \end{array}\]

Observe that the equation is in the form of a parabola, that is \(y=ax^2+bx\) where; \[a=u\tan\theta \quad;\quad b= \dfrac{g}{2u^2\cos^2\theta}\]

The equation of trajectory can also be written as

\[y = x\tan \theta \left( {1 - \frac{x}{R}} \right)\]

Here R is the horizontal range.

**What is the shape of the curve**

As y depends on the square of x, the curve must be a parabola. Therefore when any object is thrown under gravity (at some angle other than \(90^\circ\) to the horizontal plane) then it follows a parabolic path. This also implies that whenever a object moves with constant acceleration with its initial velocity not in the direction of acceleration then the particle follows parabolic path.

## A particle is thrown from ground with speed \(20m/s\) at an angle \(30^\circ\) from the horizontal. Find the horizontal range, maximum height and time of flight of the projectile?

\[\begin{array}{l} R = \dfrac{{{u^2}\sin 2\theta }}{g}\\ R = \dfrac{{{{(20)}^2}\sin {{60}^0}}}{{10}}\\ R = 20\sqrt 3 \,m \end{array}\]

\[\begin{array}{l} H = \dfrac{{{u^2}{{\sin }^2}\theta }}{{2g}}\\ H = \dfrac{{{{(20)}^2}{{\sin }^2}(30)}}{{2 \times 10}}\, = \,5m \end{array}\]

\[\begin{array}{l} T = \dfrac{{2u\sin \theta }}{g}\\ T = \dfrac{{2 \times 20\sin {{30}^0}}}{{10}} = 2\sec \end{array}\]

## The equation of trajectory of a projectile is \(y = 16x - \frac{{5{x^2}}}{4}\). Find the horizontal range.

Equation of projectile motion is \(y = x\tan \theta \left[ {1 - \frac{x}{R}} \right]\)

Given equation can be rewritten as \(y = 16x\left[ {1 - \frac{x}{{64/5}}} \right]\)

By comparing above equations \(R = \frac{{64}}{5}=12.8m\)

In a cricket match, a pace bowler \(2\text{ m}\) tall throws a yorker ball with a horizontal velocity of \(20\text{ m/s}\). The batsmen hits the ball with twice the horizontal velocity thrown by the bowler at an angle of \({45}^\circ\) with the horizontal. The ball lands just in front of a fielder who kicked the ball with his foot at an angle of \({30}^\circ\) with the horizontal in such a way that it landed just in front of the foot of the bowler. Find the velocity with which the fielder kicked the ball.

**Details and assumptions**:

- Assume the air friction to be negligible.
- Assume that the position of the batsman, bowler and fielder are collinear and the batsman hits the ball in the collinear line.
- Assume that the bowler throws the ball by stretching his hand just above the top of his head.
- Assume that the bowler does not move fom his position after releasing the ball from the crease line.
- Take \(g\) (acceleration due to gravity) as \(10{\text{m/s}}^2\).
- Round your answer (in \(\text{m/s}\)) off to the nearest integer.

A ball A of mass \(1\text{kg}\) is thrown at an angle of \({45}^\circ\) with the horizontal with kinetic energy \(50\text{J}\) such that it hits a ball B of same mass which is placed on the top of a pole. At this collision half of the kinetic energy of ball A at that instant is transferred to ball B causing ball B to move in a forward direction. If it is known that the height at which the ball is placed is the maximum height that the initial projectile of ball A would have travelled , then find the distance of the final position of ball B from the foot of the pole.

The figure below would help in understanding the situation:-

**Details and assumptions**:

- Assume that ball B does not rebound after hitting the ground.
- Assume the air friction to be negligible.
- Take \(\text{g}\) (acceleration due to gravity) as \(10\text{m}/{\text{s}}^2\).
- Give your answer(in \(\text{m}\)) to two decimal places.

## Projectile from a height

A projectile is thrown from a tower of height 'h', horizontally with speed 'u'.

Horizontal direction is taken as positive and vertical downward direction is taken as negative. The components of initial velocity and acceleration along x-axis and y-axis will be

Initial velocity along x axis \({u_x} = u\)

Initial velocity along y axis \({u_y} = 0\)

Acceleration along x axis\( = 0\)

Acceleration along y axis \(= +g\)

**Time of flight**

For motion in y direction
\[{S_y} = {u_y}t + \frac{1}{2}{a_y}{t^2}\]

Finally, the particle lands on ground and travels 'h' distance (equal to the height of tower) in vertically downward direction. Therefore displacement in y direction \({S_y} = h\)

\[\begin{align}
h& = \dfrac{1}{2}g{T^2}\\
T &= \sqrt {\dfrac{{2h}}{g}}
\end{align}\]

**Horizontal Range**

For displacement in horizontal direction
\[\begin{align}
{S_x} &= {u_x}t + \dfrac{1}{2}{a_x}{t^2}\\
R &= uT\\

R &= u\sqrt {\dfrac{{2h}}{g}}
\end{align}\]

## A ball rolls off top of a staircase with a horizontal velocity u m/s. If the steps are 'h' meter high and 'b' meter wide, the ball will just hit the edge of nth step. Then find the value of n?

If the ball hits the n-th step then displacement in vertical downward direction will be 'nh' and displacement in horizontal direction will be 'nb' \[\begin{align} R &= u\sqrt {\dfrac{{2h}}{g}}\\ nb &= u\sqrt {\dfrac{{2nh}}{g}} \\ n &= \dfrac{{2{u^2}h}}{{{b^2}g}} \end{align}\]

## A body is projected horizontally from the top of a tower with initial velocity 18 m/s. It hits the ground at angle \(45^o\). Find its vertical component of velocity when it strikes the ground?

As the projectile hits the ground at an angle \({45^o}\), thus

\[\tan {45^o}\, = \frac{{{v_y}}}{{{v_x}}}\]

The vertical and horizontal component of velocities are equal and the horizontal component of velocity always remain equals, therefore \({v_x} = 18 m/s\) \[{v_y} = {v_x} = 18 m/s\]

Two bullets of equal mass are fired horizontally with speeds of 10 m/s and 20 m/s from the same height at the same time. Which bullet will hit the ground first?

Neglect air resistance, the curvature of the earth, lift effects on the bullet, etc.

A ball of mass \(2 \text{kg}\) is placed in front of a spring having a constant of \(20 \text{N/m}\) and compression \(20 \text{m}\) at a height of \(100 \text{m}\) above the ground. Now, the spring is released, giving the ball a horizontal velocity. Find the total displacement (in metres) of the ball when it hits the ground.

**Details and assumptions**

- Assume that the ball does not rebound after hitting the ground first.
- Assume that the air friction is negligible.
- Assume that the spring is \(100\%\) efficient.
- Take \(\text{g}\) (Acceleration due to gravity) as \(10 {\text{m/s}}^2\).

## See Also

**Cite as:**Projectile Motion.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/projectile-motion-easy/