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Question: A particle is projected from a point O with a velocity u in a direction making an angle α upward wit...

A particle is projected from a point O with a velocity u in a direction making an angle α upward with the horizontal. After some time at point P it is moving at right angle with its initial direction of projection. The time of flight from O to P is

A

usin⁡αg\frac{u\sin\alpha}{g}

B

ucosecαg\frac{u\text{cosec}\alpha}{g}

C

utan⁡αg\frac{u\tan\alpha}{g}

D

usec⁡αg\frac{u\sec\alpha}{g}

Answer

ucosecαg\frac{u\text{cosec}\alpha}{g}

Explanation

Solution

When body projected with initial velocity u→\overset{\rightarrow}{u}by making angle α\alphawith the horizontal. Then after time t, (at point P) it’s direction is perpendicular tou→\overset{\rightarrow}{u}.

Magnitude of velocity at point P is given by v=ucot⁡α.v = u\cot\alpha. (from sample problem no. 9)

(

For vertical motion : Initial

velocity (at point O)=usin⁡α= u\sin\alpha

Final velocity (at point P) =−vcos⁡α=−ucot⁡αcos⁡α= - v\cos\alpha = - u\cot\alpha\cos\alpha

Time of flight (from point O to P) = t

Applying first equation of motion v=u−gtv = u - gt

−ucot⁡αcos⁡α=usin⁡α−gt- u\cot\alpha\cos\alpha = u\sin\alpha - gt

∴ t=usin⁡α+ucot⁡αcos⁡αg=ugsin⁡α[sin⁡2α+cos⁡2α]t = \frac{u\sin\alpha + u\cot\alpha\cos\alpha}{g} = \frac{u}{g\sin\alpha}\left\lbrack \sin^{2}\alpha + \cos^{2}\alpha \right\rbrack

=ucosecαg= \frac{u\text{cosec}\alpha}{g}