TOPIC-4
Basic concept
- Any particle which is thrown into space or air such that it moves under the influence of an external force (e.g. gravity, electric forces etc.) is called a projectile. The motion of such a particle is referred to as projectile motion.
it is an example of two dimensional motion with constant acceleration.
- If the force acting on the projectile is constant, then acceleration is constant. When the force is in oblique direction with the direction of initial velocity, the resultant path is parabolic.
Parabolic motion = Vertical motion + Horizontal motion
- Projectile motion can be considered to be two simultaneous motions in mutually perpendicular directions which are completely independent of each other i.e. horizontal motion and vertical motion
Ground to ground projectile
Consider a projectile thrown from horizontal ground with a
velocity u making an angle 0 with the horizontal. Take the point
of projection as origin O and the path of the projectile in the first quadrant of xy - plane, as shown in the figure. The initial velocity u is resolved in the horizontal and vertical directions i.e.
\({u_x} = u\cos \theta \) \({u_y} = u\sin \theta \)
Since gravity is the only force acting on the projectile in vertically downward direction, (ignoring air resistance)
\({a_x} = 0\)
\({a_y} = - g\)
Analyzing the motion of the projectile in horizontal and vertical directions:
Horizontal direction:
Initial velocity : \({u_x} = u\cos \theta \)
Acceleration : \({a_x} = 0\)
Velocity after time t : \({v_x} = u\cos \theta \)
Vertical direction:
Initial velocity : \({u_y} = u\sin \theta \)
Acceleration : \({a_y} = - g\)
Velocity after time t : \({v_y} = u\sin \theta - gt\)
- The position vector of the projectile after time t is \(\vec r = x\hat i + y\hat j = \left( {u\cos \theta \cdot t} \right)\hat i + \left( {u\sin \theta \cdot t - g{t^2}} \right)\hat j\);
- Velocity after time t is \(\vec v = {v_x}\hat i + {v_y}\hat j = \left( {u\cos \theta } \right)\hat i + \left( {u\sin \theta - gt} \right)\hat j\);
- Acceleration is constant ,\(\vec a = {a_x}\hat i + {a_y}\hat j = - g\hat j\)
Trajectory equation:
\(x = u\cos \theta \cdot t\)
For displacement in the vertical direction ,\(y = {u_y} \cdot t - \frac{1}{2}g{t^2}\)
\(y = u\sin \theta \cdot t - \frac{1}{2}g{t^2}\)
Substituting the value of t from eqn. (1) into eqn. (2), we get
\(y = u\sin \theta \cdot \frac{x}{{u\cos \theta }} - \frac{1}{2}g{\left( {\frac{x}{{u\cos \theta }}} \right)^2}\)
\(y = x\tan \theta - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}\)
\(y = x\tan \theta \left[ {1 - \frac{x}{R}} \right]\)
(R is the horizontal range covered by the projectile)
The equation of trajectory of the projectile is that of a parabola because the projectile covers a parabolic path.
Time of flight:
The displacement along vertical direction is zero for ground to ground projectile.
\(\left( {u\sin \theta } \right)T - \frac{1}{2}g{t^2} = 0\)
\(T = \frac{{2u\sin \theta }}{g}\)
Horizontal range:
The horizontal displacement of the projectile from the point of projection to the point it strikes the ground is called the horizontal range of the projectile.
\(R = {u_x} \cdot T\)
\(R = u\cos \theta \cdot \frac{{2u\sin \theta }}{g}\)
\(R = \frac{{{u^2}\sin 2\theta }}{g}\)
Maximum height:
Applying \({v^2} = {u^2} + 2as\) in the vertical direction between the point of projection and the topmost point, we get \({0^2} = {u^2}{\sin ^2}\theta - 2gH\)
\(H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\)
Resultant velocity, at any instant t:
\(\vec v = {v_x}\hat i + {v_y}\hat j = \left( {u\cos \theta } \right)\hat i + \left( {u\sin \theta - gt} \right)\hat j\)
\(\left| {\vec v} \right| = \sqrt {{u^2}{{\cos }^2}\theta + {{\left( {u\sin \theta - gt} \right)}^2}} \)
\(\tan \alpha = \frac{{{v_y}}}{{{v_x}}} = \frac{{u\sin \theta - gt}}{{u\cos \theta }}\)
\(\alpha \) is the angle made by the velocity vector of the projectile with the horizontal at any time instant t
General result
For maximum range \(\theta = {45^ \circ }\) \({R_{\max }} = \frac{{{u^2}}}{g}\)
In this situation \({H_{\max }} = \frac{{{u^2}{{\sin }^2}45}}{{2g}} = \frac{{{u^2}}}{{4g}}\)
\({H_{\max }} = \frac{{{R_{\max }}}}{4}\)
We get the same range for two angles of projection \(\alpha \) and ( \({90 - \alpha }\ ).But in each of the two cases, maximum height attained by the particle is different.
\(R = \frac{{2{u^2}\sin \alpha \cos \alpha }}{g} = \frac{{2{u^2}\sin \left( {90 - \alpha } \right)\cos \left( {90 - \alpha } \right)}}{g}\)
If R = H i.e. \(\frac{{{u^2}\sin 2\theta }}{g} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\)
\(\frac{{2{u^2}\sin \theta \cos \theta }}{g} = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\)
\(\therefore \tan \theta = 4\);\(\theta = {\tan ^{ - 1}}4\)
Range can also be expressed as \(R = \frac{{{u^2}\sin 2\theta }}{g} = \frac{{2u\sin \theta \cdot u\cos \theta }}{g} = \frac{{2{u_x}{u_y}}}{g}\)
Change in momentum
Initial velocity \({{\vec u}_i} = u\cos \theta \hat i + u\sin \theta \hat j\)
Final velocity \({{\vec u}_f} = u\cos \theta \hat i - u\sin \theta \hat j\)
Change in velocity from the point of projection to the point where the projectile strikes the ground.
\(\Delta \vec u = {{\vec u}_f} - {{\vec u}_i} = - 2u\sin \theta \hat j\)
Change in momentum from the point of projection to the point where the projectile strikes the ground.
\(\Delta \vec P = {{\vec P}_f} - {{\vec P}_i} = m\left( {{{\vec u}_f} - {{\vec u}_i}} \right) = m\left( { - 2u\sin \theta } \right)\hat j = - 2mu\sin \theta \hat j\)
where m is the mass of the projectile
Velocity at the highest point of the projectile is \(u\cos \theta \hat i\). Change in momentum from the point of projection to the highest point =\( - mu\sin \theta \hat j\)