Projectile Motion

Projection from a height at an angle \(\theta \) above horizontal:

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

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

Along Verticle direction \(Y\):
\({u_y} = u\sin \theta \)
\({a_y} = - g\)

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

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

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

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Equation of Trajectory:
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\(x = ut\cos \theta \)
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, \(y = x\tan \theta - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}\)

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Velocity at any instant \(t\):
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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, \({v_y} = u\sin \theta - gt\)

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

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Velocity after falling a height \(h\):
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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, \(v_y^2 = \left( {u\sin \theta } \right) - \frac{1}{2}g\left( { - h} \right)\)
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, \(\left| {\vec v} \right| = \sqrt {{u^2} + 2gh} \)

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Maximum Height attains \(H\):
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At maximum height vertical component of velocity becomes zero. At this instant \(y\) coordinate is, its maximum height.
\(v_y^2 = u_y^2 + 2{a_y}y\)
or, \(v_y^2 = {\left( {u\sin \theta } \right)^2} - 2gh\)
or, \(0 = {\left( {u\sin \theta } \right)^2} - 2gh\)
or, \(h = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\)

Therefore, the maximum height from the ground
\( = \left( {H + h} \right)\)
\( = H + \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\)

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

After solving the above equation we get the result