Question

A particle is projected from a point O with a velocity u in a direction making an angle $\alpha$ 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. $\frac{usin\alpha }{g}$
• B. $\frac{ucosec\alpha }{g}$
• C. $\frac{utan\alpha }{g}$
• D. $\frac{usec\alpha }{g}$

Answer 1

The correct option is B

$\frac{ucosec\alpha }{g}$

When body projected with initial velocity $\overrightarrow{u}$ by making angle $\alpha$ with the horizontal. Then after time t, (at point P) it’s direction is perpendicular to $\overrightarrow{u}$

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

For vertical motion : Initial velocity (at point O) $=usin\alpha$

Final velocity (at point P) $=-vcos\alpha =-ucot\alpha cos\alpha$

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

Applying first equation of motion $v=u-gt$

$-ucot\alpha cos\alpha =usin\alpha -gt\therefore t=\frac{usin\alpha +ucot\alpha cos\alpha }{g}=\frac{u}{gsin\alpha }[si{n}^2\alpha +co{s}^2\alpha ]=\frac{ucosec\alpha }{g}$