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Q.

A particle is projected from a point O with a velocity u at an angle α (upwards) to the horizontal. At a certain point P, it moves at right angles to its initial direction. It follows that

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a

OP makes an angle tan-1 (u/2g) to the horizontal

b

the distance of P from O is u/ (2g sin α)

c

the time of flight from O to P is u/(g sin α)

d

the velocity of the particle at P is u cot α

answer is C.

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Detailed Solution

(d) vcos⁡(90−α)=ucos⁡α                v=ucot⁡α (c) tan⁡(−(90−α))=usin⁡α−gtucos⁡α⇒ t=ugsin⁡α (b) r=(ucos⁡α)ugsin⁡αi^+usin⁡αugsin⁡α−12gugsin⁡α2j^r=u2gcot⁡αi^+u2g−u22gcosec2⁡αj^r=u2gcot⁡αi^+u2g1−cosec2⁡α2j^OP=u2gcot⁡α2+u2g1−cosec2⁡α22 (a) tan⁡θ=u2g1−cosec2⁡α2u22cot⁡α
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