91Ó°ÊÓ

When a particle is thrown at an angle near the earth's surface, it takes a curved path. This type of particle is known as a projectile, and its motion is known as projectile motion.

When a projectile is projected with some initial velocity \(u\) at some angle \(\theta \), the initial velocity is resolved into two components.

The horizontal component of the initial velocity \({u_x} = u\cos \theta = {v_x}\) remains constant throughout the projectile motion since no force acts on the object along the horizontal direction.

The vertical component of the initial velocity \({u_y} = u\sin \theta \) varies due to the action of the gravitational force as,

\(\begin{array}{c}{v_y} = {u_y} - gt\\ = u\sin \theta - gt\end{array}\)

Here \(g\) is the acceleration due to gravity, and \(t\) is the time.

Therefore, the velocity of the projectile at any instant of time,

\(\begin{array}{c}v = \sqrt {v_x^2 + v_y^2} \\ = \sqrt {{{\left( {u\cos \theta } \right)}^2} + {{\left( {u\sin \theta - gt} \right)}^2}} \\ = \sqrt {{u^2} - 2ugt\sin \theta + {{\left( {gt} \right)}^2}} \end{array}\)

Hence, the velocity of the projectile is never zero.

As the projectile rises up, the vertical component of the velocity decreases. At maximum height, the vertical component of the velocity becomes zero; i.e. \({v_y} = 0\), however, the horizontal component of the velocity remains constant. Therefore, the net velocity of the projectile at the maximum height is,

\(\begin{array}{c}{v_h} = \sqrt {v_x^2 + v_y^2} \\ = \sqrt {{{\left( {u\cos \theta } \right)}^2} + {{\left( 0 \right)}^2}} \\ = u\cos \theta \end{array}\)

Hence, the projectile has a minimum velocity at the maximum height.

The velocity of the projectile at the time of projection, i.e., at \(t = 0\) is,

\(\begin{array}{c}{v_{t = 0}} = \sqrt {{u^2} - 2ug \times \left( 0 \right) \times \sin \theta + {{\left[ {g \times \left( 0 \right)} \right]}^2}} \\ = u\end{array}\)

The time of flight is,

\(t = \frac{{2u\sin \theta }}{g}\)

Thus, the velocity of the projectile just before it touches the ground is,

\[\begin{array}{c}{v_f} = \sqrt {{u^2} - 2ugt\sin \theta + {{\left( {gt} \right)}^2}} \\ = \sqrt {{u^2} - 2ug\sin \theta \times \left( {\frac{{2u\sin \theta }}{g}} \right) + {{\left[ {g \times \left( {\frac{{2u\sin \theta }}{g}} \right)} \right]}^2}} \\ = u\end{array}\]

Hence, the projectile has a maximum velocity at the point of projection and the instant when it reaches the ground level.

Since the velocity is a vector quantity, the two velocities are said to be the same when the magnitude and the direction of velocity are the same. But during the projectile motion, the direction of the projectile continuously changes.

Hence, the velocity of the projectile can never have the same values as that of the initial velocity.

The velocity of the projectile at any instant of the flight is,

\(v = \sqrt {{u^2} - 2ugt\sin \theta + \left( {g{t^2}} \right)} \)

At the time of projection, i.e., at \(t = 0\), the velocity of the projectile is,

\[\begin{array}{c}v = \sqrt {{u^2} - 2ug \times \left( 0 \right) \times \sin \theta + {{\left[ {g \times \left( 0 \right)} \right]}^2}} \\ = u\end{array}\]

The time of flight is,

\(t = \frac{{2u\sin \theta }}{g}\)

Thus, the velocity of the projectile just before it touches the ground is,

\[\begin{array}{c}v = \sqrt {{u^2} - 2ug\sin \theta \times \left( {\frac{{2u\sin \theta }}{g}} \right) + {{\left[ {g \times \left( {\frac{{2u\sin \theta }}{g}} \right)} \right]}^2}} \\ = u\end{array}\]

Hence, when the projectile reaches the ground level, it has the same value of initial speed.