Two particles are simultaneously thrown from the top of two towers as shown. Their velocities are 2 m s - 1 and 14 m s - 1 . Horizontal and vertical separations between these particles are 22 m and 9 m respectively. Then the minimum separation between the particles in the process of their motion in meters is ( g = 10 m s - 2 )

Step-by-step solution

v x = 8 2 m   s - 1 = relative velocity along x-axis ∴ x = 22 - 8 2 t v y = 6 2 m / s = relative velocity along the y-axis ∴ y = 9 - 6 2 t r = x 2 + y 2 For minimum r , d r d t = 0 ⇒ t = 23 10 2   s   r = x 2 + y 2 substituting the value of t in r ,   r min = 6   m