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One second after projection, a stone moves at an angle \[45^\circ \] with horizontal. Two seconds after projection, it moves horizontally, its angle of projection is [\[g = 10\,{\text{m/}}{{\text{s}}^2}\]]
A. \[{\tan ^{ - 1}}\left( {\sqrt 3 } \right)\]
B. \[{\tan ^{ - 1}}\left( 4 \right)\]
C. \[{\tan ^{ - 1}}\left( 3 \right)\]
D. \[{\tan ^{ - 1}}\left( 2 \right)\]

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Last updated date: 24th Feb 2024
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IVSAT 2024
Answer
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Hint: Use the formula for the kinematic equation for final velocity of the object. This equation gives the relation between the final velocity, initial velocity, acceleration and time. Determine the position of the stone two times from the given information. Derive the equation for the vertical component of initial velocity of the stone for time second. Then using the given angle of stone with the horizontal, derive an expression for velocity of stone one second after projection. Hence, determine the angle of projection of the stone.

Formula used:
The first kinematic equation is given by
\[v = u + at\] …… (1)
Here, \[v\] is final velocity of the object, \[u\] is initial velocity of the object, \[a\] is acceleration of the object and \[t\] is the time.

Complete step by step answer:
We have given that the angle of the stone one second after the projection is \[45^\circ \] with the horizontal and two seconds after the projection, the stone moves horizontally.We have asked to determine the angle of projection of the stone.Let us rewrite the equation (1) for the velocity of an object in free fall.
\[{v_y} = {u_y} - gt\] …… (2)
Here, \[{v_y}\] is vertical component of final velocity, \[{u_y}\] is vertical component of initial velocity and \[g\] is acceleration due to gravity.

We have given that two seconds after projection, the stone moves horizontally. Hence, the stone must be at the top most point of the trajectory.At the top most point of projectile motion, the vertical component of velocity of the stone is zero and the vertical component of initial velocity of the stone is \[u\sin \theta \].
\[{v_y} = 0\,{\text{m/s}}\]
\[{u_y} = u\sin \theta \]
Here, \[u\] is the velocity of projection and \[\theta \] is angle of projection of the stone.

Substitute \[0\,{\text{m/s}}\] for \[{v_y}\], \[u\sin \theta \] for \[{u_y}\] and \[2\,{\text{s}}\] for \[t\] in equation (2).
\[\left( {0\,{\text{m/s}}} \right) = \left( {u\sin \theta } \right) - g\left( {2\,{\text{s}}} \right)\]
\[ \Rightarrow u\sin \theta - 2g = 0\]
\[ \Rightarrow u\sin \theta = 2g\] …… (3)
One second after the projection of stone, the angle made by the stone with the horizontal is \[45^\circ \]. Thus, the horizontal and vertical components of velocity at this point of trajectory should be the same.
Hence, the vertical component of velocity of the stone one second after the projection is \[u\cos \theta \].
\[{v_y} = u\cos \theta \]

Substitute \[u\cos \theta \] for \[{v_y}\], \[u\sin \theta \] for \[{u_y}\] and \[1\,{\text{s}}\] for \[t\] in equation (2).
\[\left( {u\cos \theta } \right) = \left( {u\sin \theta } \right) - g\left( {1\,{\text{s}}} \right)\]
\[ \Rightarrow u\cos \theta = u\sin \theta - g\]
Substitute \[2g\] for \[u\sin \theta \] in the above equation.
\[ \Rightarrow u\cos \theta = 2g - g\]
\[ \Rightarrow u\cos \theta = g\] …… (4)
Divide equation (3) by equation (4).
\[ \Rightarrow \dfrac{{u\sin \theta }}{{u\cos \theta }} = \dfrac{{2g}}{g}\]
\[ \Rightarrow \tan \theta = 2\]
\[ \therefore \theta = {\tan ^{ - 1}}\left( 2 \right)\]
Therefore, the angle of projection of the stone is \[{\tan ^{ - 1}}\left( 2 \right)\].

Hence, the correct option is D.

Note:The students should not get confused between the different values of components of velocity of the stone at different times. One second after projection, the vertical and horizontal component of the stone is the same. Hence, we have considered the horizontal component of velocity as vertical component of velocity one second after the projection of stone as value of both the components is the same.
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