Projectile Motion

Projectile motion is the curved path an object follows when it is thrown or launched into the air and moves under the influence of gravity alone (ignoring air resistance). It is a type of two-dimensional motion with horizontal and vertical components that act independently. The horizontal motion occurs at a constant speed, while the vertical motion is affected by gravity, causing the object to accelerate downward. This combination creates a parabolic trajectory. Examples include a ball being thrown, a cannonball fired, or a basketball shot at a hoop.

1.0Definition of Projectile Motion

Any object that is given an initial velocity obliquely, and that subsequently follows a path determined by the net constant force, (In this chapter constant force is gravitational force) acting on it is called a projectile.

Examples of projectile motion :

• A cricket ball hit by the batsman for a six.
• A bullet fired from a gun.
• A packet dropped from a plane; but the motion of the aeroplane itself is not projectile motion because there are forces other than gravity acting on it due to the thrust of its engine.

Assumptions of Projectile Motion:

• We shall consider only trajectories that are of sufficiently short range so that the gravitational force can be considered constant in both magnitude and direction.
• All effects of air resistance will be ignored.
• As range is very short compared to the radius of the earth, the part of the earth can be considered to be flat.

Projectile Motion:

• The motion of a projectile is known as projectile motion.
• It is an example of two-dimensional motion with constant acceleration.
• Projectile motion is considered as a combination of two simultaneous motions in mutually perpendicular directions which are completely independent from each other i.e. horizontal motion and vertical motion.

Parabolic Path = Vertical motion + Horizontal Motion

2.0Projectile Thrown At An Angle With Horizontal

• Consider a projectile thrown with a velocity u making an angle $\theta$ with the horizontal.
• Initial velocity u is resolved in components in a coordinate system in which horizontal direction is taken as x-axis,vertical direction as y-axis and point of projection as origin.

$u_x=u\cos \theta ,u_y=u\sin \theta$

• Again this projectile motion can be considered as the combination of horizontal and vertical motion.

Resultant Velocity

${\vec{V}}_R=(u\cos \theta )\hat{i}+(u\sin \theta -gt)\hat{j}$

$\left|{\vec{V}}_R\right|=\sqrt{u^2{\cos }^2\theta +(u\sin \theta -gt{)}^2}$

$\tan \alpha =\frac{u\sin \theta -gt}{u\cos \theta }$

Where $\alpha$ is the angle that velocity vector makes with horizontal, also known as direction or angle of motion.

Vectorial treatment 

Let's say a particle is projected at an angle $\theta$ from horizontal with a velocity. Now if we take the point of projection as origin and take vertically upward as positive y-axis and horizontal direction as x-axis.

$\begin{array}{rl}& \vec{u}=u\cos \theta \hat{i}+u\sin \theta \hat{j}\\ & \vec{a}=-g\hat{j}\end{array}$

Now since acceleration is uniform

Velocity after time t

$\begin{array}{rl}& \vec{V}=\vec{u}+\overrightarrow{at}\\ & \vec{V}=u\cos \theta \hat{i}+u\sin \theta \hat{j}+(-g\hat{j})t\\ & \vec{V}=u\cos \theta \hat{i}+(u\sin \theta -gt)\hat{j}\end{array}$

Displacement after time t

$\begin{array}{rl}& \vec{s}=\overrightarrow{ut}+\frac{1}{2}\overrightarrow{a{t}^2}\\ & \vec{s}=(u\cos \theta \hat{i})t+(u\sin \theta \hat{j})t+\frac{1}{2}(-g\hat{j})t^2\\ & \vec{s}=(u\cos \theta t)\hat{i}+\left(u\sin \theta -\frac{1}{2}g{t}^2\right)\hat{j}\end{array}$

Time of Flight

The displacement along the vertical direction is zero for the complete flight. Hence, along vertical direction net displacement = 0

$\Rightarrow (u\sin \theta )T-\frac{1}{2}g{T}^2=0\Rightarrow T=\frac{2u\sin \theta }{g}$

Horizontal Range

$\begin{array}{rl}& R=u_{x}T\Rightarrow R=u\cos \theta \cdot \frac{2u\sin \theta }{g}\\ & R=\frac{u^2\sin 2\theta }{g}\end{array}$

Maximum Height

At the highest point of its trajectory, the particle moves horizontally, and hence the vertical component of velocity is zero.

By using Third equation of motion

$v^2=u^2+2as$

For vertical direction

$\begin{array}{rl}& 0=u^2{\sin }^2\theta -2gH\\ & \Rightarrow H=\frac{u^2{\sin }^2\theta }{2g}\end{array}$

General Result:

• For Maximum Range $\theta =45^{\circ }$
• $R_{\max }=\frac{u^2}{g},H_{\max }=\frac{u^2}{2g}\Rightarrow H_{\max }=\frac{R_{\max }}{2}$
• We get the same range for two angle of projections $\alpha$ and (90°- $\alpha$) but in both cases, maximum heights attained by the particles are different.
• This is because, $R=\frac{u^2\sin 2\theta }{g}$ and $\sin 2\left(90^{\circ }-\alpha \right)=\sin \left(180^{\circ }-2\alpha \right)=\sin 2\alpha$
• If R=H, i.e. $\frac{u^2\sin 2\theta }{g}=\frac{u^2{\sin }^2\theta }{2g}\Rightarrow \tan \theta =4$
• Range can also be expressed as $R=\frac{u^2\sin 2\theta }{g}=\frac{2u\sin \theta \cdot u\cos \theta }{g}=\frac{2u_x{u}_y}{g}$

Equation of Trajectory

• The path followed by a particle (here projectile) during its motion is called its Trajectory. Equation of trajectory is the relation between instantaneous coordinates (Here x and y coordinate) of the particle.

If we consider the horizontal direction,

$x=u_{x}t$

$x=u\cos \theta .......(1)$

For Vertical direction

$y=u_y\cdot t-\frac{1}{2}g{t}^2$

$y=u\sin \theta \cdot t-\frac{1}{2}g{t}^2.......(2)$

Eliminating 't' from equation (1) and (2)

$y=u\sin \theta \cdot \frac{x}{u\cos \theta }-\frac{1}{2}g{\left(\frac{x}{u\cos \theta }\right)}^2\Rightarrow y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

This is an equation of parabola called the trajectory equation of projectile motion.

Other Forms of Trajectory Equation:

$y=x\tan \theta -\frac{g{x}^2\left(1+{\tan }^2\theta \right)}{2u^2}\left(\because y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }\right)$

and $y=x\tan \theta \left[1-\frac{gx}{2u^2{\cos }^2\theta \tan \theta }\right]\Rightarrow y=x\tan \theta \left[1-\frac{gx}{2u^2\sin \theta \cos \theta }\right]$

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

3.0Projectile Thrown Parallel to the Horizontal from Some Height

Consider a projectile thrown from point O at some height h from the ground with a velocity u. Now we shall study the characteristics of projectile motion by resolving the motion along horizontal and vertical directions.

Time of Flight: This is equal to the time taken by the projectile to return to ground.

From equation of motion,

$S=ut+\frac{1}{2}a{t}^2$, along vertical direction, we get

$-h=u_{y}t+\frac{1}{2}(-g)t^2\Rightarrow h=\frac{1}{2}g{t}^2\Rightarrow t=\sqrt{\frac{2h}{g}}$

Horizontal Range: Distance covered by the projectile along the horizontal direction between the point of projection to the point on the ground.

$\begin{array}{rl}& R=u_{x}t\\ & \Rightarrow R=u\sqrt{\frac{2h}{g}}\end{array}$

Velocity at a general point P(x,y)

$v=\sqrt{v_x^2+v_y^2}$

Here horizontal velocity of the projectile after time t, $v_x=u$

Velocity of projectile in vertical direction after time t, $v_y=0+(-g)t=-gt=gt(downward)$

$\therefore v=\sqrt{u^2+g^2{t}^2}and\tan \theta =\frac{v_y}{v_x}$

Velocity with Which the Projectile Hits the Ground

$\begin{array}{rl}& v_x=u\\ & v_y^2=0^2-2g(-h)\\ & v_y=\sqrt{2gh}\\ & v=\sqrt{v_x^2+v_y^2}\\ & v=\sqrt{u^2+2gh}\end{array}$

Trajectory Equation

The path traced by the projectile is called the trajectory.

After time t,

$\begin{array}{rl}& x=ut\dots \dots .(1)\\ & y=-\frac{1}{2}g{t}^2.......(2)\end{array}$

Put the value of t in equation (2)

$y=-\frac{1}{2}g\cdot \frac{x^2}{u^2}$

This is the trajectory equation of the particle projected horizontally from some height.

4.0Projection from a Tower

Case(1): Horizontal Projection

$u_x=u;u_y=0;a_y=-g$

Case (2): Projection at an angle $\theta$ above horizontal

$u_x=u\cos \theta ,u_y=u\sin \theta ,a_y=-g$

Equations of motion between A and B (in Y direction)

$\begin{array}{rl}& S_y=-h,u_y=u\sin \theta ,a_y=-g,t=T\\ & S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2\Rightarrow -h=u\sin \theta t-\frac{1}{2}g{t}^2\end{array}$

Solving this equation,we will get time of flight T

And Range, $R=u_{x}T=u\cos \theta T$

also,

$v_y^2=u_y^2+2a_y{S}_y=u^2{\sin }^2\theta +2gh;v_x=u\cos \theta$

$v_B=\sqrt{v_y^2+v_x^2}\Rightarrow v_B=\sqrt{u^2+2gh}$

Case (3): Projection at an angle   below horizontal 

$\begin{array}{rl}& u_x=u\cos \theta ;u_y=-u\sin \theta ;a_y=-g\\ & S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2\\ & S_y=-h,u_y=-u\sin \theta ;a_y=-g\\ & \Rightarrow -h=-u\sin \theta T-\frac{1}{2}g{T}^2\Rightarrow h=u\sin \theta T+\frac{1}{2}g{T}^2\end{array}$

Solving this equation,we will get time of flight T

And Range,

$\begin{array}{rl}& R=u_{x}T=u\cos \theta T\\ & v_x=u\cos \theta \\ & v_y^2=u_y^2+2a_y{S}_y=u^2{\sin }^2\theta +2(-g)(-h)\\ & v_y^2=u^2{\sin }^2\theta +2gh\\ & v_B=\sqrt{v_x^2+v_y^2}=\sqrt{u^2+2gh}\end{array}$

5.0
Projection on an Inclined Plane

Case (1): Particle is projected up the incline

Here $\alpha$ is angle of projection w.r.t. the inclined plane. The x and y axis are taken along and perpendicular to the incline as shown in the diagram.

$\begin{array}{rl}& a_x=-g\sin \beta \\ & u_x=u\cos \alpha \\ & a_y=-g\cos \beta \\ & u_y=u\sin \alpha \end{array}$

Time of Flight (T): When the particles strikes the inclined plane y becomes zero

$\begin{array}{rl}& y=u_{y}t+\frac{1}{2}{a}_y{T}^2\\ & \Rightarrow 0=u\sin \alpha T-\frac{1}{2}g\cos \beta T^2\\ & \Rightarrow T=\frac{2u\sin \alpha }{g\cos \beta }=\frac{2u_{\perp }}{g_{\perp }}\end{array}$

Where $u_{\perp }$ and $g_{\perp }$ are components of u and g perpendicular to the incline

Maximum height (H): When half of the time is elapsed, y coordinate is equal to maximum distance from the inclined plane of the projectile.

$\begin{array}{rl}& H=u\sin \alpha \left(\frac{u\sin \alpha }{g\cos \beta }\right)-\frac{1}{2}g\cos \beta {\left(\frac{u\sin \alpha }{g\cos \beta }\right)}^2\\ & H=\frac{u^2{\sin }^2\alpha }{2g\cos \beta }=\frac{u_{\perp }^2}{2g_{\perp }}\end{array}$

Range Along The Inclined Plane (R)

When the particles strikes the inclined plane x coordinate is equal to the range of the particle.

$\begin{array}{rl}& x=u_{x}t+\frac{1}{2}{a}_x{t}^2\\ & \begin{array}{rl}& \Rightarrow R=u\cos \alpha \left(\frac{2u\sin \alpha }{g\cos \beta }\right)-\frac{1}{2}g\sin \beta {\left(\frac{2u\sin \alpha }{g\cos \beta }\right)}^2\\ & \Rightarrow R=\frac{2u^2\sin \alpha \cos (\alpha +\beta )}{g{\cos }^2\beta }\end{array}\end{array}$

Case (2) : Particle is projected down the incline

In this case:

$\begin{array}{rl}& a_x=g\sin \beta ;u_x=u\cos \alpha \\ & a_y=-g\cos \beta \\ & u_y=u\sin \alpha \end{array}$

Time of Flight (T) : When the particle strikes the inclined y coordinate becomes zero.

$\begin{array}{rl}& y=u_y+\frac{1}{2}{a}_y{T}^2\\ & \Rightarrow 0=u\sin \alpha T-\frac{1}{2}g\cos \beta T^2\\ & \Rightarrow T=\frac{2u\sin \alpha }{g\cos \beta }=\frac{2u_{\perp }}{g_{\perp }}\end{array}$

Maximum Height (H):

When half of the time is elapsed y coordinate is equal to maximum height of the projectile.

$\begin{array}{rl}& H=u\sin \alpha \left(\frac{u\sin \alpha }{g\cos \beta }\right)-\frac{1}{2}g\cos \beta {\left(\frac{u\sin \alpha }{g\cos \beta }\right)}^2\\ & \Rightarrow H=\frac{u^2{\sin }^2\alpha }{2g\cos \beta }=\frac{u_{\perp }^2}{2g\perp }\end{array}$

Range Along The Inclined Plane (R)

When the particle strikes the inclined plane x coordinate is equal to the range of the particle.

$\begin{array}{rl}& x=u_{x}t+\frac{1}{2}{a}_x{t}^2\\ & \Rightarrow R=u\cos \alpha \left(\frac{2u\sin \alpha }{g\cos \beta }\right)+\frac{1}{2}g\sin \beta {\left(\frac{2u\sin \alpha }{g\cos \beta }\right)}^2\\ & \Rightarrow R=\frac{2u^2\sin \alpha \cos (\alpha -\beta )}{g{\cos }^2\beta }\end{array}$

Standard results for projectile motion on an inclined plane

Note: For a given speed, the direction which gives the maximum range of the projectile on an incline, bisects the angle between the incline and the vertical, for upward or downward projection.

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