91Ó°ÊÓ

The motion of a projectile is an example of motion with constant acceleration.When a particle is hurled obliquely near the earth's surface, the particle experiences a constant acceleration due to gravity.

• Initial height from where the arrow is projected\({h_0} = 1.5\,{\rm{m}}\).
• Velocity of projection\(u = 30\,{\rm{m}}/{\rm{s}}\).
• Angle of projection \(\theta = 60^\circ \).
• Time of flight \(t = 4.0\,{\rm{s}}\).

The motion of the arrow is represented as,

Figure: The motion of the arrow

The second equation of motion along the vertical direction is,

\(h - {h_0} = {u_y}t - \frac{1}{2}g{t^2}\)

Here\(h\)is the height of the cliff,\({h_0}\)is the initial height from where the arrow is projected, \({u_y}\)is the vertical component of the initial velocity,\(g\)is the acceleration due to gravity, and\(t\)is the time of flight.

The vertical component of the initial velocity is,

\({u_y} = u\sin \theta \)

Here \(\theta \) is the angle of projection.

From equations (1.1) and (1.2),

\(h - {h_0} = ut\sin \theta - \frac{1}{2}g{t^2}\)

Rearranging the above equation to get an expression for the height of the cliff.

\(h = {h_0} + ut\sin \theta - \frac{1}{2}g{t^2}\)

Substitute \(1.5{\rm{ m}}\) for \({h_0}\),\(30{\rm{ m}}/{\rm{s}}\) for \(u\), \(4.0{\rm{ s}}\) for \(t\), \(60^\circ \) for \(\theta \) , and \(9.8{\rm{ m}}/{{\rm{s}}^2}\) for \(g\),

\(\begin{array}{c}h = \left( {1.5{\rm{ m}}} \right) + \left( {30{\rm{ m}}/{\rm{s}}} \right) \times \left( {4.0{\rm{ s}}} \right) \times \sin \left( {60^\circ } \right) - \frac{1}{2} \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times {\left( {4.0{\rm{ s}}} \right)^2}\\ = 27{\rm{ m}}\end{array}\)

Hence, theheight of the cliff is\(27{\rm{ m}}\).

The third equation of the motion along the vertical direction is,

\(v_y^2 = u_y^2 - 2g\left( {H - {h_0}} \right)\)

Here \({v_y}\) is the velocity of the arrow at the maximum height \(\left( {{v_y} = 0} \right)\)and \(H\) is the maximum height attained by the arrow.

From equations (1.2) and (1.3),

\(0 = {\left( {u\sin \theta } \right)^2} - 2g\left( {H - {h_0}} \right)\)

Rearranging the above equation in order to get an expression for the maximum height attained by the arrow is,

\(H = {h_0} + \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}\)

Substitute \(1.5{\rm{ m}}\) for \({h_0}\), \(30{\rm{ m}}/{\rm{s}}\) for \(u\), \(60^\circ \) for \(\theta \) , and \(9.8{\rm{ m}}/{{\rm{s}}^2}\) for \(g\)

\(\begin{array}{c}H = \left( {1.5{\rm{ m}}} \right) + \frac{{{{\left( {30{\rm{ m}}/{\rm{s}}} \right)}^2} \times {{\sin }^2}\left( {60^\circ } \right)}}{{2 \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right)}}\\ \approx 36{\rm{ m}}\end{array}\)

Hence, the maximum height attained by the arrow is \(36{\rm{ m}}\).

The horizontal component of the initial velocity of the arrow remains constant throughout the projectile motion, as no force is acting on the projectile along the horizontal direction. Therefore,

\(\begin{array}{c}{v_x} = {u_x}\\ = u\cos \theta \end{array}\)

Substitute\(30{\rm{ m}}/{\rm{s}}\)for \(u\), \(60^\circ \) for \(\theta \),

\[\begin{array}{c}{v_y} = \left( {30{\rm{ m}}/{\rm{s}}} \right) \times \cos \left( {60^\circ } \right)\\ = 15{\rm{ m}}/s\end{array}\]

The vertical component of the velocity of the arrow is,

\({v_y} = {u_y} - gt\)

From equations (1.2) and (1.4),

\({v_y} = u\sin \theta - gt\)

Substitute \(30{\rm{ m}}/{\rm{s}}\) for \(u\), \(60^\circ \) for \(\theta \), \(4.0{\rm{ s}}\) for \(t\), and \(9.8{\rm{ m}}/{{\rm{s}}^2}\) for \(g\),

\(\begin{array}{c}{v_y} = \left( {30{\rm{ m}}/{\rm{s}}} \right) \times \sin \left( {60^\circ } \right) - \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \left( {4.0{\rm{ s}}} \right)\\ = - 13.22{\rm{ m}}/{\rm{s}}\end{array}\)

The magnitude of the velocity of the arrow just before hitting the cliff is,

\(v = \sqrt {v_x^2 + v_y^2} \)

Substitute \(15{\rm{ m}}/{\rm{s}}\) for \({v_x}\), and \( - 13.22{\rm{ m}}/{\rm{s}}\) for \({v_y}\),

\(\begin{array}{c}v = \sqrt {{{\left( {15{\rm{ m}}/{\rm{s}}} \right)}^2} + {{\left( { - 13.22{\rm{ m}}/{\rm{s}}} \right)}^2}} \\ \approx 20{\rm{ m}}/{\rm{s}}\end{array}\)

Hence, the speed of the arrow just before hitting the cliff is \(20{\rm{ m}}/{\rm{s}}\).