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# The ceiling of a hall is $40m$ high. For maximum horizontal distance, the angle at which the ball can be thrown with the speed of $56m{s^{ - 1}}$ without hitting the ceiling of the hall is (take $g = 9.8m/{s^2}$ )(A) ${25^{\rm O}}C$ (B) ${30^{\rm O}}C$ (C) ${45^{\rm O}}C$ (D) ${60^{\rm O}}C$

Last updated date: 19th Sep 2024
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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.

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.

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 ${45^{\rm O}}$.