Projectile Motion

A Projectile projected at an angle \(\theta \) with the horizontal from the ground || Ground to Ground Projection

Let a projectile be fired from point O with initial velocity \(\vec u\) such that it makes an angle \(\theta \) with the horizontal direction.
Now resolve the velocity \(\vec u\) into two components: (i) \({u_x} = u\cos \theta \) and (ii) \({u_y} = u\sin \theta \).
Here the horizontal component, \({u_x} = u\cos \theta \) remains the same throughout the journey of the projectile because no force acts on the projectile due to gravity in the horizontal direction.
On the other hand, the verticle component \({u_y} = u\sin \theta \) goes on decreasing as the projectile goes up from the point of projection and ultimately becomes zero at the heighest point A. This is because the earth attracts the projectile towards its centre. After attaining the highest point, the projectile starts its journey downwards. Now, the vertical component \({u_y}\) starts increasing and becomes equal to \(u\sin \theta \), when it hits the ground. This is because the earth is exerting a downward force on the projectile. In this case, the velocity of the body at the highest point = \(u\cos \theta \).
Thus, Projectile motion is the combined effect of horizontal motion with uniform velocity (i.e. no acceleration) and the vertical motion with uniform acceleration.

Projectile fired at an angle with the horizontal:
Let a projectile be thrown with unitial velocity \(\vec u\) at an angle \(\theta \) with the horizontal direction. Thus, horizontal component of the initial velocity \({u_x} = u\cos \theta \) and vertical component of the initial velocity \({u_y} = u\sin \theta \). The projectile travels in the horizontal direction with a constant velocity \(u\cos \theta \) and it travels in the vertical direction with a velocity \(u\sin \theta \). The vertical component of the velocity gradually reduces to zero at the highest point and then the projectile moves downward to fall onto the ground. The trajectory of the particle is represented by OAB.
Let at any intant of time \(t\), the projectile be at point P. Let \(x\) and \(y\) be the horizontal and vertical distances travelled by projectile in time \(t\).

Horizontal Motion:
\({u_x} = u\cos \theta \)
\({a_x} = 0\)
\({S_x} = R\) Verticle Motion:
\({u_y} = u\sin \theta \)
\({a_y} = - g\)
\({S_y} = H\)

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Equation of Motion:
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Here the projectile is not accelerated horizontally as no force acts on it in horizontal direction or velocity of projectile in the horizontal direction is constant. Distance travelled by the projectile in the horizontal direction in time \(t\) is given by
\({S_x} = {u_x}t + \frac{1}{2}{a_x}{t^2}\)
or, \(x = u\cos \theta .t + 0\)
or, \(x = ut\cos \theta \)
Distance travelled by the projectile in the vertical direction in time \(t\) is given by
\({S_y} = {u_y}t + \frac{1}{2}{a_y}{t^2}\)
or, \(y = u\sin \theta .t - \frac{1}{2}g{t^2}\)
or, \(y = ut\sin \theta - \frac{1}{2}g{t^2}\)

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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 \(y\)
\(y = u\left( {\frac{x}{{u\cos \theta }}} \right)\sin \theta - \frac{1}{2}g{\left( {\frac{x}{{u\cos \theta }}} \right)^2}\)
or, \(y = x\tan \theta - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }}\)
Which is the equation of a parabola of second order symmetric about \(y\) axis. Hence, the path followed by the projectile is parabolic.

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Velocity at any time t:
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A projectile is projected with a velocity \(\vec u\) at an angle \(\theta \) with the horizontal. Thus, horizontal component of the initial velocity \({u_x} = u\cos \theta \) and vertical component of the initial velocity \({u_y} = u\sin \theta \).
\({v_x} = {u_x} + {a_x}t\)
or, \({v_x} = u\cos \theta + 0\)
or, \({v_x} = u\cos \theta \)
\({v_y} = {u_y} + {a_y}t\)
or, \({v_y} = u\sin \theta - gt\)

Therefore, \(v = {v_x}\hat i + {v_y}\hat j\)
or, \(\vec v = u\cos \theta \hat i + \left( {u\sin \theta - gt} \right)\hat j\)
or, \(\left| {\vec v} \right| = \sqrt {{{\left( {u\cos \theta } \right)}^2} + {{\left( {u\sin \theta - gt} \right)}^2}} \)
or, \(\left| {\overrightarrow v } \right| = \sqrt {{u^2}{{\cos }^2}\theta + {{\left( {u\sin \theta - gt} \right)}^2}} \)
Let the velocity \(v\) makes an angle \(\beta \) with the horizontal direction than
\(\tan \beta = \frac{{{v_y}}}{{{v_x}}}\)
or, \(\tan \beta = \frac{{u\sin \theta - gt}}{{u\cos \theta }}\)
or, \(\tan \beta = \tan \theta - \frac{{gt}}{{u\cos \theta }}\)
or, \(\beta = {\tan ^{ - 1}}\left( {\tan \theta - \frac{{gt}}{{u\cos \theta }}} \right)\)

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Velocity at any height:
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\(v_x^2 = u_x^2 + 2{a_x}x\)
or, \(v_x^2 = {\left( {u\cos \theta } \right)^2} + 0\)
or, \({v_x} = u\cos \theta \)
\(v_y^2 = u_y^2 + 2{a_y}y\)
or, \(v_y^2 = {\left( {u\sin \theta } \right)^2} - 2gh\)
or, \({v_y} = \sqrt {{{\left( {u\sin \theta } \right)}^2} - 2gh} \)

Therefore, \(\overrightarrow v = {v_x}\hat i + {v_y}\hat j\)
or, \(\left| {\vec v} \right| = \sqrt {v_x^2 + v_y^2} \)
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 by the Projectile (H):
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The maximum vertical distance travelled by the projectile during its journey is called the maximum height attained by the projectile. It is denoted by H.
For vertical upward motion, \(v_y^2 = u_y^2 + 2{a_y}H\) and at maximum height \({v_y} = 0\)
Now \(v_y^2 = u_y^2 + 2{a_y}H\)
or, \(0 = {\left( {u\sin \theta } \right)^2} - 2gH\)
or, \(2gH = {u^2}{\sin ^2}\theta \)
or, \(H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\)
Thus, maximum height attained by a projectile depends upon (i) the initial velocity of the projectile and (ii) the angle of projection with the horizontal.

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Time of Flight of a Projectile:
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The total time taken by the projectile from the point of projection till it hits the horizontal plane having point of projection is called time of flight.
Now vertical distance travelled by projectile in time \(t\) is
\({S_y} = {u_y}T + \frac{1}{2}{a_y}{T^2}\)
At \(t = T = \) time of flight,
then \(y = 0\) because net displacement in \(y\) direction is zero, when the projectile hits the ground.
Thus,
or, \(0 = u\sin \theta .T - \frac{1}{2}g{T^2}\)
or, \(\frac{1}{2}g{T^2} = uT\sin \theta \)
or, \(T = \frac{{2u\sin \theta }}{g}\)

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Horizontal Range (R) of a Projectile:
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The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits is called horizontal range. Therefore, the maximum horizontal distance travelled by the projectile is called horizontal range.
For horixontal motion,
\(x = {u_x}t + \frac{1}{2}{a_x}{t^2}\)
Here, \({u_x} = u\cos \theta \) and \(t = T = \) time of flight then, \(x = R\)
\(x = {u_x}.T\)
or, \(x = u\cos \theta .\frac{{2u\sin \theta }}{g}\)
or, \(R = \frac{{{u^2}\sin 2\theta }}{g}\)