Projectile Motion

A projectile thrown at height \(H\) from the ground with a initial velocity \(u\) in horizontal direction || Horizontal projection from height.

Consider a projectile thrown from point \(O\) at some height \(H\) from the ground with a velocity \(u\) in horizontal direction.

Horizontal Motion:

\({u_x} = u\)
\({a_x} = 0\)

Vertical Motion:

\({u_y} = 0\)
\({a_y} = - g\)

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Equation of Motion or Displacement:
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Along \(X\) direction:
\({S_x} = {u_x}t + \frac{1}{2}{a_x}{t^2}\)
or, \(x = ut + 0\) \(\left[ {{a_x} = 0} \right]\)
or, \(x = ut\)

Along \(Y\) direction:
\({S_y} = {u_y}t + \frac{1}{2}{a_y}{t^2}\)
or, \(y = 0 - \frac{1}{2}g{t^2}\)
or, \(y = - \frac{1}{2}g{t^2}\)

Therefore, the displacement of the particle is expressed by
\(\vec r = x\hat i + y\hat j\)
or, \(\vec r = ut\hat i - \frac{1}{2}g{t^2}\hat j\)
or, \(\left| {\vec r} \right| = \sqrt {{{\left( {ut} \right)}^2} + {{\left( {\frac{1}{2}g{t^2}} \right)}^2}} \)

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Equation of Trajectory:
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The path traced by the projectile is called its trajectory.
\(x = ut\)
or, \(t = \frac{x}{u}\)

Putting the value of \(x\) in \(y\)
\(y = - \frac{1}{2}g{t^2}\)
or, \(y = - \frac{1}{2}g{\left( {\frac{x}{u}} \right)^2}\)
or, \(y = - \frac{{g{x^2}}}{{2{u^2}}}\)

This is equation of a parabola. The above equation is called trajectory equation.

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

Along \(Y\) direction:
\({v_y} = {u_y} + {a_y}t\)
or, \({v_y} = 0 - gt\) \(\left[ {{u_y} = 0} \right]\)

Therefore,
\(\vec v = v_x\hat i + v_y\hat j\)
or, \(\vec v = u\hat i - gt\hat j\)
or, \(\left| {\vec v} \right| = \sqrt {{{\left( u \right)}^2} + {{\left( {gt} \right)}^2}} \)
or, \(\vec v = \sqrt {{u^2} + {g^2}{t^2}} \)

\(\tan \theta = \frac{{{v_y}}}{{{v_x}}}\)
or, \(\tan \theta = - \frac{{gt}}{u}\), Here negative sign indicates clockwise direction.

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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, \(v_x^2 = {u^2} + 0\)
or, \({v_x} = u\)

Along \(Y\) direction:
\(v_y^2 = u_y^2 + 2{a_y}y\)
or, \(v_y^2 = 0 - 2g( - h)\)
or, \(v_y^2 = 2gh\)
or, \({v_y} = \sqrt {2gh} \)

Therefore,
\(\vec v = {v_x}\hat i + {v_y}\hat j\)
or, \(\vec v = u\hat i + \sqrt {2gh} \hat j\)
or, \(\left| {\vec v} \right| = \sqrt {{u^2} + 2gh} \)

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

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Time of Flight \(T\):
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Time of flight is the time for which projectile remains in air. At time \(T\), the 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, \( - H = 0 - \frac{1}{2}g{T^2}\) At heightest point \(\left[ {{u_y} = 0} \right]\) = 0
or, \(H = \frac{1}{2}g{T^2}\)
or, \(g{T^2} = 2H\)
or, \(T = \sqrt {\frac{{2H}}{g}} \)

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Range \(R\):
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Distance covered by the projectile along the horizontal direction between the point of projection to the point on the ground.
\({S_x} = {u_x}t + \frac{1}{2}{a_x}{t^2}\)
or, \(R = uT + 0\) \(\left[ {{a_x} = 0} \right]\)
or, \(R = uT\)
or, \(R = u\sqrt {\frac{{2H}}{g}} \)


\(\left| {\vec r} \right| = \sqrt {{{\left( {ut} \right)}^2} + {{\left( {\frac{1}{2}g{t^2}} \right)}^2}} \)

\(y = - \frac{{g{x^2}}}{{2{u^2}}}\)

\(\vec v = u\hat i - gt\hat j\)

\(\vec v = u\hat i + \sqrt {2gh} \hat j\)

Maximum height is \(H\)

\(T = \sqrt {\frac{{2H}}{g}} \)

\(R = u\sqrt {\frac{{2H}}{g}} \)