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 $$t(\sqrt{2} + 1)$$ s. The value of t is _____ (use $$g = 10 \ m s^{-2}$$)

given height of point B above base = 10 m
left incline = 45°, right incline = 30°, g = 10 m/s²

motion from A to B

the block is projected with just sufficient speed to reach B
so velocity at B becomes zero

using energy:

½ m u² = m g h
u² = 2gh = 2 × 10 × 10 = 200

(this part only confirms that velocity at B = 0; time is not needed here)

motion from B to C

initial velocity u = 0

acceleration along incline:

a = g sin30° = 10 × 1/2 = 5 m/s²

length of incline BC:

sin30° = height / BC
1/2 = 10 / BC
BC = 20 m

apply kinematics:

s = ut + ½at²

20 = 0 + ½ × 5 × t²
20 = (5/2)t²

t² = 8
t = 2√2

given time = t(√2 + 1)

so,

2√2 = t(√2 + 1)

t = 2√2 / (√2 + 1)

multiply numerator and denominator by (√2 − 1):

t = 2√2(√2 − 1) / (2 − 1)
t = 2(2 − √2)

which evaluates to 2