Projectile Motion

Projection from a Height at an angle \(\theta \) below horizontal.

Oblique projection from a certain height, i.e. Projection from a height at an angle \(\theta \) below horizontal.

Along Horizontal direction \(X\):
\({u_x} = u\cos \theta \)
\({a_x} = 0\)

Along Verticle Direction \(Y\):
\({u_y} = - u\sin \theta .t\)
or, \({u_y} = - ut\sin \theta \)
\({a_y} = - g\)

==================================
Displacement or Equation of Motion:
==================================
Along \(X\) direction:
\({S_x} = {u_x}t + \frac{1}{2}{u_x}{t^2}\)
or, \(x = u\cos \theta .t + 0\)
or, \(x = ut\cos \theta \)

Along \(Y\) direction:
\({S_y} = {u_y}t + \frac{1}{2}{a_y}{t^2}\)
\(y = - u\sin \theta .t - \frac{1}{2}g{t^2}\)
or, \(y = - ut\sin \theta - \frac{1}{2}g{t^2}\)

Therefore,
\(\vec r = x\hat i + y\hat j\)
or, \(\vec r = \left( {ut\cos \theta } \right)\hat i + \left( { - ut\sin \theta - \frac{1}{2}g{t^2}} \right)\hat j\)
or, \(\left| {\vec r} \right| = \sqrt {{{\left( {ut\cos \theta } \right)}^2} + {{\left( {ut\sin \theta + \frac{1}{2}g{t^2}} \right)}^2}} \)

======================
Equation of Trajectory:
======================
Here, \(x = ut\cos \theta \)
or, \(t = \frac{x}{{u\cos \theta }}\)

Putting the value of \(t\) in \(Y\), we get
\(y = - ut\sin \theta - \frac{1}{2}g{t^2}\)
or, \(y = - u\left( {\frac{x}{{u\cos \theta }}} \right)\sin \theta - \frac{1}{2}g{\left( {\frac{x}{{u\cos \theta }}} \right)^2}\)
or, \(y = - x\tan \theta - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}\)

=============================
Velocity at any instant \(t\):
=============================
Along \(X\) direction:
\({v_x} = {u_x} + {a_x}t\)
or, \({v_x} = u\cos \theta + 0\) \(\left[ {{a_x} = 0} \right]\)
or, \({v_x} = u\cos \theta \)

Along \(Y\) direction:
\({v_y} = {u_y} + {a_y}t\)
or, \({v_y} = - u\sin \theta - gt\)
or, \({v_y} = - \left( {u\sin \theta + gt} \right)\)

Therefore, Velocity at any instant \(t\)
\(\vec v = {v_x}\hat i + {v_y}\hat j\)
or, \(\vec v = u\cos \theta \hat i - \left( {u\sin \theta + gt} \right)\hat j\)
or, \(\left| {\vec v} \right| = \sqrt {{{\left( {u\cos \theta } \right)}^2} + {{\left( {u\sin \theta + gt} \right)}^2}} \)

=====================================
Velocity after falling a height \(h\):
=====================================
Along \(X\) direction:
\(v_x^2 = u_x^2 + 2{a_x}x\)
or, \(v_x^2 = {\left( {u\cos \theta } \right)^2} - 0\) \(\left[ {{a_x} = 0} \right]\)
or, \(v_x^2 = {u^2}{\cos ^2}\theta \)
or, \({v_x} = u\cos \theta \)

Along \(Y\) direction:
\(v_y^2 = u_y^2 + 2{a_y}y\)
or, \(v_y^2 = {\left( { - u\sin \theta } \right)^2} - 2g\left( { - h} \right)\)
or, \(v_y^2 = {\left( {u\sin \theta } \right)^2} + 2gh\)

Therefore, velocity after falling a height \(h\)
\(\vec v = {v_x}\hat i + {v_y}\hat j\)
or, \(\vec v = u\cos \theta \hat i + \sqrt {{{\left( {u\sin \theta } \right)}^2} + 2gh} \hat j\)
or, \(\left| {\vec v} \right| = \sqrt {{{\left( {u\cos \theta } \right)}^2} + {{\left( {u\sin \theta } \right)}^2} + 2gh} \)
or, \(\left| {\vec v} \right| = \sqrt {{u^2} + 2gh} \)

====================
Maximum Height \(H\):
====================
Here maximum height is \(H\)

===================
Time of Flight \(T\):
===================
At time \(T\) the particle will be at ground again, i.e. displacement along \(Y\) axis
\({S_y} = {u_y}t + \frac{1}{2}{a_y}{t^2}\)
\( - H = - ut\sin \theta - \frac{1}{2}g{t^2}\)
\(g{t^2} + 2ut\sin \theta - 2H = 0\)

After solving the equation, we get the result.