Two inclined planes are placed as shown in figure. A block is projected from the point A of inclined plane AB 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 BC. 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/s²)

Idea: do the two stretches separately. Both are frictionless inclines, so the acceleration along each is g sin θ, and the answer's (√2+1) tells you the two times will come out as 2√2 and 2.

A to B, on the 45° face. Length AB=(10)/(sin 45°)=10√2 m and retardation g sin 45°=(10)/(√2)=5√2 m/s².

'Just sufficient' means the block arrives at B with zero speed, so u²=2(5√2)(10√2)=200 and u=10√2 m/s.

t₁=u/(5√2)=(10√2)/(5√2)=2 s.

B to C, on the 30° face. Length BC=(10)/(sin 30°)=20 m and acceleration g sin 30°=5 m/s², starting from rest:

20=1/2(5)t₂², so t₂²=8 and t₂=2√2 s.

Total. t₁+t₂=2+2√2=2(√2+1) s.

Matching against t(√2+1) gives t=2.