
From energy conservation at point A and B,
21mv02=mgh
⇒v0=gh=10×10
⇒v0=102ms−1
For A→B
At B,v=0
Acceleration of a particle moving on a smooth incline is gsinθ. Therefore, along AB, a=−gsin45∘=2−10ms−2.
Using equation of motion,
v=u+at1
0=102−210t1⇒t1=2s
For B→C
Using second equation of motion,
s=ut2+21at22
⇒sin30∘10=21(10sin30∘)t22
t2=22s
So total time T=t1+t2 =22+2
=2(2+1)s