Q: A particle is projected at angle 60º with speed 10 √3, from the point ‘ A ‘ as shown in the fig. At the same time the wedge is made to move with speed 10 √3 towards right as shown in the figure. Then the time after which particle will strike with wedge is

(g = 10 m/sec^{2}) :

(a) 2 sec

(b) 2 √3 sec

(c) 4/ √3 sec

(d) none of these

**Click to See Solution : **

Sol: Suppose particle strikes wedge at height ‘ S ‘ after time t.

$ S = 15 t – \frac{1}{2}.10 . t^2 $

$ S = 15 t – 5 t^2 $

During this time , distance travelled by particle in horizontal direction = 5 √3 t.

Also wedge has travelled extra distance

$x = \frac{S}{tan 30^o}$

$x = \frac{15 t – 5 t^2}{\frac{1}{\sqrt{3}}}$

Total distance travelled by wedge in time t = 10 √3 t

= 5 √3 t + √3 (15 t – 5 t^{2})

t = 2 sec