Question

A, B, C, D are the points in a vertical line, the lengths AB=BC=CD . If the body Falls from rest at A , prove that the times of describing AB, BC, CD are as 1:√2-√1:√3-√2 .

Answer 1

Let AB=BC=CD= h ; Let t1 be the time taken to travel the
distance AB.
speed u1 at end of the time t1, when the body is at B is
given by

u1 = g×t1 ......................(1)
h = (1/2)×g×t1×t1..............(2)

Let t2 be the time taken to travel the distance BC.

h = u1×t2+(1/2)g×t2×t2 = g×t1×t2 + (1/2) g×t2×t2
...............(3)

Initial speed during the travel of BC is substituted using
(1) in the above relation.
Using the equations (2) and (3) and after canceling the
common factors, we get

t1² = 2t1× t2 + t2².............(4)

By solving the quadratic equation for t2, we get t2 = t1×(√2-1)..................(5)

Speed u2 after travelling(t1+t2) seconds, when the body
is at C is given by

u2 = g×(t1+t2) = gt1×√2..........(6)

​​​​​​Let t3 be the time taken to travel the distance CD.

h = u2×t3+(1/2)g×t3×t3 = g×t1×√2×t3 + (1/2)g ×t3×t3
................(7)

initial speed during the travel of CD is substituted using
(6) in the above relation.
Using the equations (2) and (7) and after canceling the
common factors, we get

t1²= 2×√2×t1+ t3².................(8)

By solving the quadratic equation for t3, we get t3 =
t1×(√3-√2)...............(9)

From (5) & (9)
So t1:t2:t3 = t1: t1×(√2-1):t1×(√3-√2)

i.e. 1:(√2-√1):(√3-√2)