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A. \[55^\circ \]

B. \[45^\circ \]

C. \[60^\circ \]

D. \[75^\circ \]

Answer

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\[v = u + at\] and find the horizontal and vertical component of velocity.

Use the formula:

\[\tan \theta = \,\,\dfrac{{{v_Y}}}{{{v_X}}}\] to find the angle.

In this given problem, the plane is flying horizontally, whose velocity is \[100\,{\text{m}}{{\text{s}}^{ - 1}}\]. During the flight, it suddenly drops an object from it. The object does not fall horizontally, rather it will move at an inclination with the horizontal.

Let the angle at which the object falls to the ground be \[\theta \].

In the figure, the initial and the final velocity of the horizontal motion are indicated along with the initial and final velocity of the vertical motion.

\[{u_X}\] indicates the initial velocity along the horizontal motion.

\[{v_X}\] indicates the final velocity along the horizontal motion.

\[{u_Y}\] indicates the initial velocity along the vertical motion.

\[{v_Y}\] indicates the final velocity along the vertical motion.

Applying the formula, along the vertical component:

\[{u_X} = 100\,{\text{m}}{{\text{s}}^{ - 1}}\], \[{v_X} = ?\], \[{a_X} = 0\] and \[t = 10\,{\text{s}}\]

\[

{v_X} = {u_X} + {a_X}t \\

= 100 + 0 \times 10 \\

= 100\,{\text{m}}{{\text{s}}^{ - 1}} \\

\]

Applying the formula, along the horizontal component:

\[{u_Y} = 0\,{\text{m}}{{\text{s}}^{ - 1}}\], \[{v_X} = ?\], \[{a_X} = 10\,{\text{m}}{{\text{s}}^{ - 2}}\] and \[t = 10\,{\text{s}}\]

\[

{v_Y} = {u_Y} + {a_Y}t \\

= 0 + 10 \times 10 \\

= 100\,{\text{m}}{{\text{s}}^{ - 1}} \\

\]

The horizontal component of velocity is \[100\,{\text{m}}{{\text{s}}^{ - 1}}\] and the vertical component of velocity is \[100\,{\text{m}}{{\text{s}}^{ - 1}}\].

To find angle at which the object hits the ground:

Find tangent:

\[

\tan \theta = \,\,\dfrac{{{v_Y}}}{{{v_X}}} \\

\tan \theta = \,\,\dfrac{{100}}{{100}} \\

\tan \theta = \,\,1 \\

\theta = \,\,{\tan ^{ - 1}}\left( 1 \right) \\

\]

\[\theta = 45^\circ \]