Question

The velocity of a projectile at any instant is $u$ making an angle $\alpha$ to the horizon. The time after which it will be moving at right angle to this direction is

• A. $\frac{ucosec\alpha }{g}$
• B. $\frac{ucos\alpha }{g}$
• C. $\frac{usin\alpha }{g}$
• D. $\frac{utan\alpha }{g}$

Answer 1

The correct option is A $\frac{ucosec\alpha }{g}$

At any timr t,the horizontal component x and the vertical component y of displacement of the projected ball with an angle a to horizontal and initial velocity u is given by:

$x=utcosa$.

$y=utsina-\left(1/2\right)g{t}^2.$

At the start the slope of the ball to horizontal =$tan\left(a\right)$....(1)

At any time t the slope of the ball $=dy/dx=\frac{\left(dy/dt\right)}{\left(dx/dt\right)}$

$=\frac{(u\ast tcosa)}{(u\ast tsina-(1/2\left)t^2\right)}$

$dy/dx=\frac{(usina-gt)}{cosa}..$.......(2)

If the direction of the ball is perpendicular to the initial direction, then

$dy/dx\ast tana=-1$. Use the value of dy/dx obtained in (2):

(($usina-gt$$/ucosa)\ast tana=-1$

$usi{n}^{2}a-gtsina=-uco{s}^{2}a$

$u(si{n}^{2}a+co{s}^{2}a)=gtsina$

$u=gtsina.$

Therefore $t=\frac{u}{gsina}=\frac{ucoseca}{g}$.