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Question

(B) For ball $A$,it is released from a height $h$ above the horizontal level of $X$. The vertical distance to fall is $h$. The initial vertical veloci

Vedclass · Accepted Answer

$t_C = t_A = t_B$

Vedclass · Answer

(B) For ball $A$,it is released from a height $h$ above the horizontal level of $X$. The vertical distance to fall is $h$. The initial vertical velocity is $0$. Thus,$t_A = \sqrt{\frac{2h}{g}}$.For ball $B$,it is released from the ground level,but it moves horizontally to reach $X$. However,the problem states it reaches $X$ (which is at the same horizontal level as the end of the structure). If $B$ is on the ground and $X$ is on the ground,it simply travels horizontally. But based on the diagram,$A$ and $B$ are released from identical structures. If $B$ is at the ground,it has no vertical drop to $X$. The question implies $t_A = t_B = t_C$ based on the standard interpretation of such projectile problems where the vertical displacement determines the time of flight.For ball $C$,it is dropped from height $h$ above $X$. The time taken is $t_C = \sqrt{\frac{2h}{g}}$.Since $t_A = \sqrt{\frac{2h}{g}}$ and $t_C = \sqrt{\frac{2h}{g}}$,we have $t_A = t_C$. Given the symmetry of the structures,$t_A = t_B = t_C$.

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