Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane A B along its surface with a velocity just sufficient to carry it to the top Point B at a height 10 m . After reaching the Point B the block slides down on inclined plane B C . Time it takes to reach to the point C from point A is t 2 + 1 s . The value of t is _____ (use g = 10 m s - 2 )

Step-by-step solution

From energy conservation at point A and B , 1 2 m v 0 2 = m g h ⇒ v 0 = g h = 10 × 10 ⇒ v 0 = 10 2   m   s - 1 For A → B At B ,    v = 0 Acceleration of a particle moving on a smooth incline is g sin θ . Therefore, along A B , a = - g sin 45 ° = - 10 2   m   s - 2 . Using equation of motion, v = u + a t 1 0 = 10 2 - 10 2 t 1 ⇒ t 1 = 2   s For B → C Using second equation of motion, s = u t 2 + 1 2 a t 2 2 ⇒ 10 sin 30 ° = 1 2 1