Question

An archer shoots an arrow with a velocity of 30m/s at an angle of 20 degrees with respect to the horizontal. An assistant standing on the level ground \[30\,{\text{m}}\] downrange from the launch point throws an apple straight up with the minimum initial speed necessary to meet the path of the arrow. What is the initial speed of the apple and at what time after the arrow is launched should the apple be thrown so that the arrow hits the apple?
A. \[0.1\,{\text{s}}\]
B. \[0.01\,{\text{s}}\]
C. \[1\,{\text{s}}\]
D. \[0.001\,{\text{s}}\]

Answer 1

Hint: Use the three kinematic equations of motion. Determine the horizontal and vertical components of initial velocity of the arrow. Also determine the initial velocity of the apple. Determine the times at which the arrow and apple meet each other and subtract them to determine the required time.

Formulae used:
The expression for the velocity \[v\] is
\[v = \dfrac{x}{t}\] …… (1)
Here, \[x\] is displacement and \[t\] is time.
The kinematic equation in terms of the vertical displacement \[y\], initial vertical velocity \[{v_y}\], acceleration \[g\] and time \[t\] is
\[y = {v_y}t - \dfrac{1}{2}g{t^2}\] …… (2)
The kinematic equation in terms of the final vertical velocity \[v\], vertical displacement \[y\], initial vertical velocity \[{v_y}\] and acceleration \[g\] is
\[{v^2} = v_y^2 - 2gy\] …… (3)
The kinematic equation in terms of the final vertical velocity \[v\], initial vertical velocity \[{v_y}\] and acceleration \[g\] and time \[t\] is
\[v = {v_y} - gt\] …… (4)

Complete step by step answer:
We can see that from the given information, the angle of projection of the arrow is \[20^\circ \] and speed of projection is \[30\,{\text{m/s}}\].
\[\theta = 20^\circ \]
\[v = 30\,{\text{m/s}}\]
The path followed by the arrow shows that the arrow is in the projectile motion with a constant horizontal velocity.
The horizontal displacement \[x\left( t \right)\] of the arrow at any time \[t\] is given by
\[x\left( t \right) = {v_x}t\]
\[ \Rightarrow x\left( t \right) = v\cos \theta t\]
Here, \[{v_x}\] is the horizontal component of velocity which is given by \[v\cos \theta \].
Substitute \[30\,{\text{m/s}}\] for \[v\] and \[20^\circ \] for \[\theta \] in the above equation.
\[ \Rightarrow x\left( t \right) = \left( {30\,{\text{m/s}}} \right)\cos 20^\circ t\]
\[ \Rightarrow x\left( t \right) = 28.2t\]
 Rewrite equation (2) for the vertical displacement \[y\left( t \right)\] of the arrow at time \[t\].
\[y\left( t \right) = {v_y}\left( t \right)t - \dfrac{{g{t^2}}}{2}\]
\[ \Rightarrow y\left( t \right) = v\sin \theta t - \dfrac{{g{t^2}}}{2}\]
Here, \[{v_y}\] is the vertical component of velocity which is given by \[v\sin \theta \].
Substitute \[30\,{\text{m/s}}\] for \[v\], \[20^\circ \] for \[\theta \] and \[9.8\,{\text{m/}}{{\text{s}}^2}\] for \[g\]in the above equation.
\[ \Rightarrow y\left( t \right) = \left( {30\,{\text{m/s}}} \right)\sin 20^\circ t - \dfrac{{\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right){t^2}}}{2}\]
\[ \Rightarrow y\left( t \right) = 10.3t - 4.9{t^2}\] …… (5)
Let us now determine the time at which the arrow is at the vertical displacement of \[30\,{\text{m}}\].
Rewrite equation (1) for the time at which arrow is at the vertical displacement of \[30\,{\text{m}}\].
\[{t_{arrow}} = \dfrac{{30\,{\text{m}}}}{{{v_x}}}\]
Substitute \[28.2\,{\text{m/s}}\] for \[{v_x}\] in the above equation.
\[{t_{arrow}} = \dfrac{{30\,{\text{m}}}}{{28.2\,{\text{m/s}}}}\]
\[ \Rightarrow {t_{arrow}} = 1.06\,{\text{s}}\]
Hence, from the time of projection, the arrow is at horizontal distance \[30\,{\text{m}}\] at time \[1.06\,{\text{s}}\].
Let us determine the vertical displacement of the arrow at time \[1.06\,{\text{s}}\].
Substitute \[1.06\,{\text{s}}\] for \[t\] in equation (5).
\[ \Rightarrow y\left( {1.06\,{\text{s}}} \right) = 10.3\left( {1.06\,{\text{s}}} \right) - 4.9{\left( {1.06\,{\text{s}}} \right)^2}\]
\[ \Rightarrow y\left( {1.06\,{\text{s}}} \right) = 5.4\,{\text{m}}\]
Now determine the minimum velocity of the apple with which it should be thrown upward.
The velocity of the apple when it reaches the maximum height is zero. Hence, the final velocity of the apple in the air is zero.
Substitute for and for in equation (3).
\[0 = v_{\min }^2 - 2gy\left( {1.06\,{\text{s}}} \right)\]
\[ \Rightarrow {v_{\min }} = \sqrt {2gy\left( {1.06\,{\text{s}}} \right)} \]
Substitute \[9.8\,{\text{m/}}{{\text{s}}^2}\] for \[g\] and \[5.4\,{\text{m}}\] for \[y\left( {1.06\,{\text{s}}} \right)\] in the above equation.
\[ \Rightarrow {v_{\min }} = \sqrt {2\left( {9.8\,{\text{m/}}{{\text{s}}^2}} \right)\left( {5.4\,{\text{m}}} \right)} \]
\[ \Rightarrow {v_{\min }} = 10.3\,{\text{m/s}}\]
Hence, the minimum initial velocity of the apple should be \[10.3\,{\text{m/s}}\].
The time when the apple is in this motion using equation (4) is
\[{t_{apple}} = \dfrac{{{v_{\min }}}}{g}\]
Substitute \[10.3\,{\text{m/s}}\] for \[{v_{\min }}\] and \[9.8\,{\text{m/}}{{\text{s}}^2}\] for \[g\] in the above equation.
\[ \Rightarrow {t_{apple}} = \dfrac{{10.3\,{\text{m/s}}}}{{9.8\,{\text{m/}}{{\text{s}}^2}}}\]
\[ \Rightarrow {t_{apple}} = 1.05\,{\text{s}}\]
The time \[t\] after which the apple should be thrown upward after the projection of the arrow is subtraction of \[{t_{arrow}}\] and \[{t_{apple}}\].
\[t = {t_{arrow}} - {t_{apple}}\]
Substitute \[1.06\,{\text{s}}\] for \[{t_{arrow}}\] and \[1.05\,{\text{s}}\] for \[{t_{apple}}\] in the above equation.
\[t = \left( {1.06\,{\text{s}}} \right) - \left( {1.05\,{\text{s}}} \right)\]
\[ \therefore t = 0.01\,{\text{s}}\]

Therefore, the apple should be thrown \[0.01\,{\text{s}}\] after the projection of arrow.Hence, the correct option is D.

Note:One can also use the equation of trajectory of the projectile to determine the time required for the arrow to reach the horizontal position at the apple is to be thrown vertically and meet with the apple.