## Q.6A particle is projected upwards with a velocity of 100 m/s at an angle of 60, with the vertical. Then time taken by the particle when it will move perpendicular to its initial direction-

A) 10sec

B) 20 by root 3

C) 5 sec

D) 10 by root 3

09-Nov-2015 11:15 AM
~2

0

Initially, angle made by horizontal = (90 - 60) = 30°

When the projectile is making 90° with the initial, then angle made by the projectile with horizontal = 60°.

Since there is no acceleration in horizontal, horizontal velocity component will remain same:

ucos(30) = vcos(60)

=> v = 100$\dpi{80} \small \sqrt{3}$

Considering vertical components of velocities in the equation v = u + at:

=> -vsin60 = usin30 + (-10)t

and using v = 100$\dpi{80} \small \sqrt{3}$, we get the time t = 20 secs

( this is not matchng with the answers given, so I guess the options given might be misprinted or it could be the question. Although, if we take the initial angle to be 60° with the horizontal, the we get ans t = 20/$\dpi{80} \small \sqrt{3}$ )

09-11-2015 16:24
~4