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(B) The time of flight of a projectile depends only on the vertical component of its initial velocity and the acceleration due to gravity.In this prob

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$\frac{2V \cos 	heta}{g}$

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(B) The time of flight of a projectile depends only on the vertical component of its initial velocity and the acceleration due to gravity.In this problem,the ball is projected at an angle $	heta$ with the vertical. Therefore,the vertical component of the initial velocity is $V_y = V \cos 	heta$.The time of flight $T$ is given by the formula $T = \frac{2 V_y}{g} = \frac{2 V \cos 	heta}{g}$.Since the collision with the vertical wall is elastic,the horizontal component of the velocity changes direction but its magnitude remains the same. The vertical component of the velocity remains completely unaffected by the collision with a vertical wall.Thus,the total time of flight remains the same as it would have been without the wall,which is $T = \frac{2 V \cos 	heta}{g}$.

$A$ ball is projected from the ground with a velocity $V$ at an angle $\theta$ to the vertical. On its path,it makes an elastic collision with a vertical wall and returns to the ground. The total time of flight of the ball is:

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