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(C) Let the velocity of the plane be $\vec{v}_{P} = u_{0} \hat{i}$.When the bomb is dropped,its initial velocity is the same as that of the plane,so $

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straight line vertically down from the plane

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(C) Let the velocity of the plane be $\vec{v}_{P} = u_{0} \hat{i}$.When the bomb is dropped,its initial velocity is the same as that of the plane,so $\vec{v}_{B, initial} = u_{0} \hat{i}$.At any time $t$,the velocity of the bomb relative to the ground is $\vec{v}_{B} = u_{0} \hat{i} - gt \hat{j}$.The velocity of the plane at any time $t$ is $\vec{v}_{P} = u_{0} \hat{i}$.The velocity of the bomb relative to the observer in the plane is $\vec{v}_{B/P} = \vec{v}_{B} - \vec{v}_{P} = (u_{0} \hat{i} - gt \hat{j}) - u_{0} \hat{i} = -gt \hat{j}$.Since the relative velocity is always directed vertically downwards,the trajectory of the bomb as seen by an observer in the plane is a straight line vertically downwards.

$A$ bomb is dropped by a fighter plane flying horizontally. To an observer sitting in the plane,the trajectory of the bomb is a:

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