Projectile

Any object projected into space or air, such that it moves under the effect of gravity alone, is called a projectile. It may be pointed out that for throwing the projectile into space, an engine or a fuel of some kind may be used but after its firing, the projectile should move under the effect of gravity alone. Thus, airplane in flight is not a projectile; but say a bomb drop from the airplane is the projectile. Other examples of projectiles are

(i) a bullet fired from the gun
(ii) a shell fight from a canon
(iii) A javelin or hammer thrown by an athlete,
(iv) I stone thrown from the top of Hill or a tower, etc.

The path followed by the projectile is called its trajectory.

A projectile moving in earth atmosphere is acted upon by the resistance of air and the shape of the trajectory is greatly affected by air resistance. Except for very light projectiles, the air resistance is negligible. While driving expression for the time of flight, horizontal range, etc. for a projectile, the air friction will be neglected. In the absence of air resistance, the motion of the projectile is considered as the combination of the following two independent motions:

1. Motion along the horizontal path with uniform velocity and
2. Motion along the vertical path under gravity  i.e. with uniform acceleration equal to g.

Projectile fired parallel to horizontal

Consider that a projectile is fired from the point O at a certain height above the ground with velocity u parallel to the horizontal. Suppose that A is the point on the ground vertically below the point of projection O. Regard the point O as the origin and horizontal line OX (in the direction of the velocity of projection of the projectile) as the positive direction of the x-axis and a vertical line OY in upward direction positive direction of the y-axis. Obviously, at t = 0, the position of the projectile along the x-axis and y-axis is given by

$ \displaystyle {{x}_{0}}=0,\text{ }{{y}_{0}}=0$

As the projectile moves, it covers distance along the horizontal due to the velocity of projection and along the vertical in a downward direction due to the velocity, it attains under the effect of gravity. Suppose that at any time t, the projectile reaches point P, so that the distances covered along X and Y-axis are x and -y respectively.

Motion along horizontal

If we neglect friction due to air, then velocity of the projectile along the horizontal will remain constant. Thus,

$ \displaystyle \begin{array}{l}\text{initial velocity along orizontal, }{{u}_{x}}=u;\\\text{acceleration along horizontal, }{{a}_{x}}=0\end{array}$

The position of the projectile at any time t along the X-axis is given by

$ \displaystyle {x={{x}_{0}}+{{u}_{x}}t+\frac{1}{2}{{a}_{x}}{{t}^{2}}}$

$ \displaystyle {\text{Setting }{{x}_{0}}=0,}$ $ \displaystyle {{{u}_{x}}=u,}$ and $ \displaystyle {{{a}_{x}}=0,}$ we have

$ \displaystyle {x=0+ut+\dfrac{1}{2}\left( 0 \right){{t}^{2}}}$

$ \displaystyle {\Rightarrow x=ut}$

$ \displaystyle {\Rightarrow t=\dfrac{x}{u}}$

Motion along vertical

Since, the projectile is fired parallel to the horizontal, its initial velocity along the vertical, $ \displaystyle {{u}_{x}}=0$.

Also, acceleration along the vertical, $ \displaystyle {{a}_{y}}=-g$

The negative sign is due to the reason that acceleration due to gravity in downward direction along negative Y-axis.

The position of the projectile along Y-axis at any time t is given by

$ \displaystyle {y={{y}_{0}}+{{u}_{y}}t+\frac{1}{2}{{a}_{y}}{{t}^{2}}}$

$ \displaystyle {\text{Setting }{{y}_{0}}=0,\text{ }{{u}_{y}}=0,}$ and $ \displaystyle {{{a}_{y}}=-g\text{ and }y=-y,}$ we have

$ \displaystyle {-y=0+0\times t+\frac{1}{2}\left( {-g} \right){{t}^{2}}}$

$ \displaystyle {y=\frac{1}{2}g{{t}^{2}}}$

Substituting for $ \displaystyle t$in the above equation, we have

$ \displaystyle {y=\frac{1}{2}g{{{\left( {\frac{x}{u}} \right)}}^{2}}}$

$ \displaystyle {y=\left( {\frac{g}{{2{{u}^{2}}}}} \right){{x}^{2}}}$

It is the equation of a parabola, which is symmetrical about Y-axis. Hence, a projectile fired along the horizontal moves along a parabolic path.

Projectile fired at an angle with horizontal

Consider that a project file is fired with velocity you and making an angle theta with the horizontal from the point O on the ground. Regard the point O as the origin, horizontal line OX as the positive direction of the x-axis and the vertical line oy in upward direction as a positive direction of y-axis. Suppose that at t = 0, the projectile is at the point O. Obviously the positions of the projectile along x-axis and y-axis are

$ \displaystyle {{x}_{0}}=0,\text{ }{{y}_{0}}=0$

The projectile move, it covers distance along the horizontal due to the horizontal component $ \displaystyle u\cos \theta $ of the velocity of projection and along the vertical due to the vertical component $ \displaystyle u\sin \theta $. Suppose that at any time t, the projectile reaches the point P so that its positions along the x and y axis are given by x and y respectively.

Motion along horizontal

If we neglect friction due to air, then horizontal component of the velocity i.e. $ \displaystyle u\cos \theta $ will remain constant. Thus,

Initialvelocity along horizontal, $ \displaystyle {{u}_{x}}=u\cos \theta $; acceleration along the horizontal,  $ \displaystyle {{a}_{x}}=0$.

The position of the projectile along X-axis at any time t is given by

$ \displaystyle {x={{x}_{0}}+{{u}_{x}}t+\frac{1}{2}a{{t}^{2}}}$

$ \displaystyle {\text{Setting }{{x}_{0}}=0,}$ $ \displaystyle {{{u}_{x}}=u\cos \theta }$ and $ \displaystyle {{{a}_{x}}=0,\text{ we have}}$

$ \displaystyle {x=0+\left( {u\cos \theta } \right)t+\frac{1}{2}\left( 0 \right){{t}^{2}}}$

$ \displaystyle {\Rightarrow x=\left( {u\cos \theta } \right)t}$

$ \displaystyle {\Rightarrow t=\frac{x}{{u\cos \theta }}}$

Motion along vertical

The velocity of the projectile along vertical goes on decreasing due to the effect of gravity.

Initial velocity along vertical, $ \displaystyle {{u}_{y}}=u\sin \theta $; acceleration along vertical, $ \displaystyle {{a}_{y}}=-g$

The position of the projectile along y-axis at any time t is given by

$ \displaystyle {y={{y}_{0}}+{{u}_{y}}t+\frac{1}{2}{{a}_{y}}{{t}^{2}}}$

$ \displaystyle {\text{Setting }{{y}_{0}}=0,}$ $ \displaystyle {{{u}_{y}}=u\sin \theta }$ and $ \displaystyle {{{a}_{y}}=-g,\text{ we have}}$

$ \displaystyle {y=\left( {u\sin \theta } \right)t+\frac{1}{2}\left( {-g} \right){{t}^{2}}}$

$ \displaystyle {\Rightarrow y=\left( {u\sin \theta } \right)t-\frac{1}{2}g{{t}^{2}}}$

Equation of Trajectory

The equation of the trajectory of the projectile can be obtained by substituting the value of $ \displaystyle t=\frac{x}{{u\cos \theta }}$ in the above equation. Therefore, we have

$ \displaystyle \begin{array}{l}y=\left( {u\sin \theta } \right)\dfrac{x}{{u\cos \theta }}-\dfrac{1}{2}g{{\left( {\dfrac{x}{{u\cos \theta }}} \right)}^{2}}\\\Rightarrow y=x\tan \theta -\left( {\dfrac{g}{{2{{u}^{2}}{{{\cos }}^{2}}\theta }}} \right){{x}^{2}}\end{array}$

As, the above equation is the equation of a parabola, it follows that a projectile fires at some angle with the horizontal moves along a parabolic path.

Time of flight

It is the time taken by the projectile to return to ground or the time for which the projectile remains in air above the horizontal plane from the point of projection. It is noted by T.

As the motion from the point O to A and then from A to O are symmetrical, the time of ascent (for journey from point O to A) and the time of descent (for journey from A to O) will be each equal to T/2. Further, on reaching the highest point A, the vertical component of the velocity of the projectile must become zero i.e. $ \displaystyle {{v}_{y}}=0$.

The velocity of the projectile at any time t along Y-axis is given by

$ \displaystyle {{{v}_{y}}={{u}_{y}}+{{a}_{y}}t}$

$ \displaystyle {\text{Setting }{{u}_{y}}=u\sin \theta }$; $ \displaystyle {{{a}_{y}}=-g;\text{ }t=\frac{T}{2}}$ and $ \displaystyle {{{v}_{y}}=0\text{, we have}}$

$ \displaystyle {0=u\sin \theta +\left( {-g} \right)\frac{T}{2}}$

$ \displaystyle {\Rightarrow T=\frac{{2u\sin \theta }}{g}}$

Maximum height attained

It is the greatest height to which a projectile rises above the point of projection. It is denoted by H.

The distance covered by the projectile in the time t along the y-axis is given by

$ \displaystyle {y={{y}_{0}}+{{u}_{y}}t+\frac{1}{2}{{a}_{y}}{{t}^{2}}}$

$ \displaystyle {\text{Setting }{{y}_{0}}=0,\text{ }{{u}_{y}}=u\sin \theta }$ and $ \displaystyle {{{a}_{y}}=-g,\text{ }t=\frac{T}{2}=\frac{{u\sin \theta }}{g}}$ and $ \displaystyle {y=H,\text{ we have}}$

$ \displaystyle H=\left( {u\sin \theta } \right)\times \frac{{u\sin \theta }}{g}$ $ \displaystyle +\frac{1}{2}\left( {-g} \right){{\left( {\frac{{u\sin \theta }}{g}} \right)}^{2}}$

$ \displaystyle {H=\frac{{{{u}^{2}}{{{\sin }}^{2}}\theta }}{{2g}}}$

Horizontal range

It is the distance covered by projectile along horizontal between the point of projection to the point on ground, where the projectile returns again. It is denoted by R.

Obviously, the horizontal range are is the horizontal distance covered by the projectile with uniform velocity $ \displaystyle {u\cos \theta }$ in a time equal to the time of flight. Therefore,

$ \displaystyle \begin{array}{l}R=u\cos \theta \times T\\=u\cos \theta \times \dfrac{{2u\sin \theta }}{g}\\=\dfrac{{{{u}^{2}}\left( {2\sin \theta \cos \theta } \right)}}{g}\\\because 2\sin \theta \cos \theta =\sin 2\theta ,\text{ we have}\\R=\dfrac{{{{u}^{2}}\sin 2\theta }}{g}\end{array}$

(i) Angle of projection for maximum range: For a given velocity of projection, the horizontal range will be maximum, when

$ \displaystyle \sin 2\theta =1;\text{ or }2\theta ={{90}^{\circ }}\text{ or }\theta ={{45}^{\circ }}$

(ii) Maximum horizontal range: The expression for the maximum horizontal range can be obtained by setting $ \displaystyle \sin 2\theta =1$ in the equation $ \displaystyle R=\dfrac{{{{u}^{2}}\sin 2\theta }}{g}$

$ \displaystyle {{R}_{{\max }}}=\frac{{{{u}^{2}}}}{g}$

(iii) Horizontal range is same for angle of projection $ \displaystyle \theta \text{ and }\left( {{{{90}}^{\circ }}-\theta } \right)$: For the angle of projection $ \displaystyle \theta $, the horizontal range is given by

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

Let $ \displaystyle {{R}’}$ be the horizontal range for the angle of projection $ \displaystyle \left( {{{{90}}^{\circ }}-\theta } \right)$. Then,

$ \displaystyle \begin{array}{l}{R}’=\dfrac{{{{u}^{2}}\sin 2\left( {{{{90}}^{\circ }}-\theta } \right)}}{g}\\=\dfrac{{{{u}^{2}}\sin \left( {{{{180}}^{\circ }}-2\theta } \right)}}{g}\\\because \sin \left( {{{{180}}^{\circ }}-2\theta } \right)=\sin 2\theta ,\text{ we have}\\{R}’=\dfrac{{{{u}^{2}}\sin 2\theta }}{g}\end{array}$

Thanks the horizontal range is same for angle of projection $ \displaystyle \theta \text{ and }\left( {{{{90}}^{\circ }}-\theta } \right)$. When angle of projection is $ \displaystyle \left( {{{{90}}^{\circ }}-\theta } \right)$, the velocity of projection will make angle $ \displaystyle \theta $ with vertical and horizontal range is same in two cases.