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**Hint:**We know that using the third equation of motion \[{V^2} - {U^2} = 2as\] we can write the given relation as \[{V^2}{\sin ^2}\theta - 2gH = 0\]. We will put the value of V, g, H in the equation and find the value of the angle for the maximum horizontal distance.

**Complete step by step answer:**

It is given in the question that the ceiling of a hall is \[40m\] high. Then we have to find the angle at which the ball can be thrown with the speed \[56m{s^{ - 1}}\] so that it will cover the maximum horizontal distance.

We know that according to the third equation of motion \[{V^2} - {U^2} = 2as\]. Here the initial velocity of the ball is considered as \[0m{s^{ - 1}}\] and the distance S is given by H and acceleration is given by g. then we can rewrite this equation as \[{V^2}{\sin ^2}\theta - 2gH = 0\].

\[56 \times 56{\sin ^2}\theta - 2 \times 9.8 \times 40 = 0\]

\[56 \times 56{\sin ^2}\theta = 2 \times 9.8 \times 40\]

\[{\sin ^2}\theta = \dfrac{{2 \times 9.8 \times 40}}{{56 \times 56}}\]

\[{\sin ^2}\theta = \dfrac{{784}}{{3136}}\]

\[{\sin ^2}\theta = 0.25\]

On finding the square root of both the sides we get-

\[\sin \theta = 0.50\]

We know that the value of \[\sin \theta = 0.50\] an angle \[{30^{\rm O}}\].

Therefore, the maximum value \[\theta \] will be \[{30^{\rm O}}\].

**Thus, option B is correct.**

**Additional information:**

It is Note:d that in an open field we want to cover the maximum distance then we have to project our ball at an angle of \[{45^{\rm O}}\]. We will get the maximum range \[{45^{\rm O}}\].

**Note:**

It is Note:d that the majority one does mistake in writing the relation of maximum height, they may merge the relation of range and height, to avoid such mistake we can use these equations of motion to derive the exact relation and solve our problem accordingly.

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