An arrow is shot from a height of \(1.5{\rm{ m}}\) toward a cliff of height \(H\). It is shot with a velocity of \(30{\rm{ m}}/{\rm{s}}\) at an angle of \(60^\circ \) above the horizontal. It lands on the top edge of the cliff \(4.0{\rm{ s}}\) later.

(a) What is the height of the cliff?

(b) What is the maximum height reached by the arrow along its trajectory?

(c) What is the arrow's impact speed just before hitting the cliff?

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 \(20{\rm{ m}}/{\rm{s}}\).