A) \[u\tan \alpha \]
B) \[u\cot \alpha \]
C) \[u\operatorname{cosec}\alpha \]
D) \[u\sec \alpha \]
Correct Answer: B
Solution :
[b] Horizontal velocity at point '[O'=u cos\[\alpha \]Horizontal velocity at point 'P'=v sin \[\alpha \] In projectile motion horizontal component of velocity remains constant throughout the motion \[\therefore v\sin \alpha =u\cos \alpha \] \[\Rightarrow v=u\cot \alpha \]You need to login to perform this action.
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