Projectile Motion For Pre-Medical Entrance Examination: NEET

When a body is projected such that velocity of projection is not parallel to the force, then it moves along a curved path. This motion is called two dimensional motion. If force on the body is constant then curved path of the body is parabolic. This motion is studies under projectile motion.
🔴 It is an example of two dimensional motion.
🔴 It is an example of motion with constant (or uniform) acceleration. Thus equation of motion can be used to analyze projectile motion.
🔴 A particle thrown in the space which moves under the effect of gravity only is called only a "projectile". The motion of this particle is referred to as projectile motion.
🔴 If a particle possesses a uniform acceleration in a directions oblique to its initial velocity, the resultant path will be parabolic. Let X-axis is along the ground and Y-axis the vertical, then the path of projectile projected at an angle $\theta$ from the ground is as shown.

GROUND TO GROUND PROJECTION:

Projectile motion can be considered as two mutually perpendicular motions, which are independent of each other.
Projectile Motion = Horizontal Motion + Vertical Motion

Horizontal Motion:
🔴 Equation of Motion along Horizontal Direction:
🔷 Initial Velocity in horizontal direction $= {u_x} = u\cos \theta$
🔷 Acceleration along horizontal direction ${a_x} = 0$ [Neglecting air resistance]
🔷 Therefore, horizontal velocity remains unchanged.
🔷 At any instant horizontal velocity ${u_x} = u\cos \theta$
At any time, $t$ X co-ordinate or displacement along X-direction is $x = {u_x}t = u\cos \theta$

Vertical Motion:
🔴 Equation of Motion along Vertical Direction:
It is motion under the effect of gravity so that as particle moves upwards the magnitude of its vertical velocity decreases.
🔷 Initial velocity in vertical direction ${u_y} = u\sin \theta$
🔷 Acceleration along vertical direction ${a_y} = - g$
🔷 At any time $t$, displacement in vertical direction or "height" of the particle above the ground $y = {u_y}t - \frac{1}{2}g{t^2} = u\sin \theta - \frac{1}{2}g{t^2}$

🔴 Resultant Motion:
Net initial velocity $= \vec u = {u_x}\hat i + {u_y}\hat j = u\cos \theta \hat i + u\sin \hat j$
Direction of $u$ can be explained in terms of angle $\theta$ it makes with the ground.
Net Acceleration $\vec a = {a_x}\hat i + {a_y}\hat j = - g\hat j$ [direction of $g$ is downwards]

🔴 Co-ordinate of the particle at time $t$:
Co-ordinate of the particle $\left( {x,y} \right) = \left( {{u_x}t,{u_y}t} \right) = \left( {ut\cos \theta ,ut\sin \theta - \frac{1}{2}g{t^2}} \right)$
Resultant displacement of the particle in time $t$, $\sqrt {{x^2} + {y^2}}$
or, Resultant displacement = $= \sqrt {{{\left( {ut\cos \theta } \right)}^2} + \left( {ut\sin \theta - \frac{1}{2}g{t^2}} \right)}$

🔴 Velocity of Particle at any time $t$
Velocity of particle at any time $t$ is $\vec v = {v_x}\hat i + {v_y}\hat j = {u_x}\hat i + \left( {{u_y} - gt} \right)\hat j$
$= u\cos \theta \hat i + \left( {u\sin \theta - gt} \right)\hat j$

The magnitude of velocity at time $t$ $= \left| {\vec v} \right| = \sqrt {v_x^2 + v_y^2} = \sqrt {{{\left( {u\cos \theta } \right)}^2} + {{\left( {u\sin \theta - gt} \right)}^2}}$

If angle made by velocity $\vec v$ with the ground is $\alpha$ then,
$\tan \alpha = \frac{{{v_y}}}{{{v_x}}} = \frac{{{u_y} - gt}}{{{u_x}}} = \frac{{u\sin \theta - gt}}{{u\cos \theta }}$
or, $\tan \alpha = \tan \theta - \frac{{gt}}{{u\cos \theta }}$

🔴 Change in Velocity and Momentum of Projectile:

When a particle returns to ground again at point B, its co-ordinate is zero and the magnitude of its velocity is $u$ at angle $\theta$ with the ground.

The total angular change is $= 2\theta$
Now, The initial velocity of the projectile ${\vec u_i} = u\cos \theta \hat i + u\sin \theta \hat j$ And
The Final velocity of the projectile ${\vec u_f} = u\cos \theta \hat i - u\sin \theta \hat j$
Therefore, The change in velocity $= \left( {{{\vec u}_i} - {{\vec u}_f}} \right) = \Delta \vec u$
And $\left| {\Delta \vec u} \right| = 2u\sin \theta$

The change of momentum is $\left| {\Delta \vec p} \right| = m\left| {\Delta \vec v} \right| = 2mu\sin \theta$

🔴 The time of flight $\left( T \right)$ of Projectile:
Time of flight is the time for which projectile remains in air.
Suppose at time $\left( T \right)$ the particle will be at ground again, i.e. displacement along Y-axis becomes zero.
Since, $y = {u_y}t - \frac{1}{2}g{t^2}$
or, $0 = ut\sin \theta - \frac{1}{2}g{t^2}$
or, $g{t^2} = 2ut\sin \theta$
or, ${t^2} = \frac{{2ut\sin \theta }}{g}$
or, $t = \frac{{2u\sin \theta }}{g}$
Here total time of flight $T = \frac{{2u\sin \theta }}{g}$

We know Time of ascent = Time of descent = $\frac{T}{2}$ = $\frac{{{u_y}}}{g} = \frac{{u\sin \theta }}{g}$
At time $\frac{T}{2}$, the particle attains the maximum height of its trajectory.

🔴 Maximum Height attained $H$:
At maximum height vertical component of velocity becomes zero. At this instant Y-co-ordinate is, its maximum height.
Since, $v_y^2 = u_y^2 - 2gy$
or, ${0^2} = u_y^2 - 2gH$ [${v_y} = 0$, $y = H$ ]
or, $H = \frac{{u_y^2}}{{2g}} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$

🔴 Horizontal Range or Range $\left( R \right)$
It is the displacement of particle along X-direction during its complete flight.
Since, $x = {u_x}t$
Therefore, Range $R = {u_x}T = {u_x}\frac{{2{u_y}}}{g}$
or, Range $R = \frac{{2{u_x}{u_y}}}{g}$
or, $R = \frac{{2\left( {u\cos \theta } \right)\left( {u\sin \theta } \right)}}{g} = \frac{{{u^2}\left( {2\sin \theta \cos \theta } \right)}}{g} = \frac{{{u^2}\sin 2\theta }}{g}$

🔴 Maximum Horizontal Range ${R_{\max }}$:
If value of $\theta$ is increased from $\theta = 0^\circ$ to $\theta = 90^\circ$, then range increases from $\theta = 0^\circ$ to $\theta = 45^\circ$, but it decreases beyond $45^\circ$.
For maximum Range, $\theta = 45^\circ$ and ${R_{\max }} = \frac{{{u^2}\sin 2\left( {45^\circ } \right)}}{g} = \frac{{{u^2}}}{g}$
Therefore, ${R_{\max }} = \frac{{{u^2}}}{g}$

🔴 Comparison of two projectiles of equal range:

When two projectile are thrown with equal speeds at angles $\theta$ and $(90 - \theta )$ then, their ranges equal but maximum heights attained are different and time of flights are also different.
At Angle $\theta$,
$R = \frac{{{u^2}\sin 2\theta }}{g}$ And,
At Angle, $\left( {90 - \theta } \right)$
${R^1} = \frac{{{u^2}\sin 2\left( {90 - \theta } \right)}}{g} = \frac{{{u^2}\sin \left( {180 - 2\theta } \right)}}{g} = \frac{{{u^2}\sin 2\theta }}{g}$

Therefore, $R = {R^1}$

Now maximum height of projectiles,
$H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ And
${H^1} = \frac{{{u^2}{{{\mathop{\rm Sin}\nolimits} }^2}\left( {90 - \theta } \right)}}{{2g}} = \frac{{{u^2}{{\cos }^2}\theta }}{{2g}}$

Now,
🔷 $\frac{H}{{{H^1}}} = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = {\tan ^2}\theta$
🔷 $H{H^1} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}} \times \frac{{{u^2}{{\cos }^2}\theta }}{{2g}} = \frac{{{u^4}{{\sin }^2}\theta {{\cos }^2}\theta }}{{4{g^2}}} = \frac{{{u^4}{{\left( {2\sin \theta \cos \theta } \right)}^2}}}{{16g}} = \frac{{{R^2}}}{{16}}$
🔷 $H + {H^1} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}} + \frac{{{u^2}{{\cos }^2}\theta }}{{2g}} = \frac{{{u^2}}}{{2g}}$

TIME OF FLIGHT OF PROJECTILE:
$T = \frac{{2{u_y}}}{g} = \frac{{2u\sin \theta }}{g}$ And
${T^1} = \frac{{2{u_y}}}{g} = \frac{{2u\sin \left( {90 - \theta } \right)}}{g} = \frac{{2u\cos \theta }}{g}$

🔷 $\frac{T}{{{T^1}}} = \frac{{\frac{{2u\sin \theta }}{g}}}{{\frac{{2u\cos \theta }}{g}}} = \frac{{\sin \theta }}{{\cos \theta }} = \tan \theta$
🔷 $T{T^1} = \frac{{2u\sin \theta }}{g}\frac{{2u\cos \theta }}{g} = \frac{{4{u^2}\sin \theta \cos \theta }}{{{g^2}}} = \frac{{2{u^2}\sin 2\theta }}{{{g^2}}} = \frac{{2R}}{g}$
or, $T{T^1} \propto R$

🔴 EQUATION OF TRAJECTORY:
Equation of motion along horizontal direction, $x = {u_y}t = ut\cos \theta$
Equation of motion along vertical direction, $y = {u_y}t - \frac{1}{2}g{t^2} = ut\sin \theta - \frac{1}{2}g{t^2}$

Eliminating $t$ from these two equation,
$t = \frac{x}{{u\cos \theta }}$ ... ... ... (i)
Thus, $y = u\left( {\frac{x}{{u\cos \theta }}} \right)\sin \theta - \frac{1}{2}g{\left( {\frac{x}{{u\cos \theta }}} \right)^2}$
or, $y = x\tan \theta - \frac{1}{2}g\frac{{{x^2}}}{{{u^2}{{\cos }^2}\theta }}$

This is an equation of a parabola, so it can be stated that projectile always follows a parabolic path.
$y = x\tan \theta \left( {1 - \frac{{gx}}{{2{u^2}\sin \theta \cos \theta }}} \right) = x\tan \theta \left( {1 - \frac{x}{R}} \right)$

🔴 KINETIC ENERGY OF A PROJECTILE:
Kinetic energy $= \frac{1}{2} \times \left( {mass} \right) \times {\left( {velocity} \right)^2}$
Let a body is projected with a velocity $u$ at an angle $\theta$. Thus the initial kinetic energy of projectile
${K_0} = \frac{1}{2}m{u^2}$

Since velocity of projectile at maximum height is $u\cos \theta$, then kinetic energy at highest point,
$K = \frac{1}{2} \times m \times {\left( {u\cos \theta } \right)^2} = \frac{1}{2}m{u^2}{\cos ^2}\theta = {K_0}{\cos ^2}\theta$
This is the minimum kinetic energy during whole motion.


🔴🔴SOME TRICKS:🔴🔴
🔷 (1) At maximum height ${V_y} = 0$ and, ${V_x} = {u_x} = u\cos \theta$
So, At maximum height $V = \sqrt {V_x^2 + V_y^2}$
or, $V = u\cos \theta$


🔷 (2) At maximum height angle between velocity and acceleration is $90^\circ$


🔷 (3) Magnitude of velocity at height $h$ is:
Initial vertical velocity ${u_y} = u\sin \theta$
Final vertical velocity is ${v_y}$
Acceleration is: $- g$

$v_y^2 = u_y^2 - 2gh$
or, $v_y^2 = {\left( {u\sin \theta } \right)^2} - 2gh$
or, ${v_y} = \sqrt {{{\left( {u\sin \theta } \right)}^2} - 2gh}$

And Initial horizontal velocity = final horizontal velocity = ${v_x} = {u_x} = u\cos \theta$
Therefore, $v = \sqrt {v_x^2 + v_y^2} = \sqrt {{{\left( {u\cos \theta } \right)}^2} + {{\left( {u\sin \theta } \right)}^2} - 2gh}$
or, $v = \sqrt {{u^2} - 2gh}$


🔷 (4) Here time of flight of projectile: $T = \frac{{2{u_y}}}{g} = \frac{{2u\sin \theta }}{g}$
Total height attained by the projectile: $H = \frac{{u_y^2}}{{2g}} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
Total horizontal Range: $R = \frac{{2{u_x}{u_{_y}}}}{g} = \frac{{2u\cos \theta \times u\sin \theta }}{g} = \frac{{{u^2}\sin 2\theta }}{g}$

Here $T$ and $H$ depends only upon $u$ and angle of projection $\theta$.

If two projectile thrown in different direction, have equal time of flight $T$, then their vertical speeds are same, so that their maximum height are is also same.
If ${H_A} = {H_B}$, then ${\left( {{u_y}} \right)_A} = {\left( {{u_y}} \right)_B}$ and also ${T_A} = {T_B}$


🔷 (5) If $\theta = 45^\circ$, then
${R_{\max }} = \frac{{{u^2}}}{g}$ and,
$H = \frac{{{u^2}}}{{4g}}$
Hence, ${R_{\max }} = 4H$


🔷 (6) When $R = H$
$\frac{{2{u_x}{u_y}}}{g} = \frac{{u_y^2}}{{2g}}$
$\frac{{2{u^2}\sin \theta \cos \theta }}{g} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$
or, $\frac{R}{H} = 4\cot \theta = 1$
or, $\cot \theta = \frac{1}{4}$
or, $\tan \theta = 4$
or, $\theta = {\tan ^{ - 1}}\left( 4 \right) = 76^\circ$