Let us assume
PR = 2x
So, PQ = QR = x
The ball begins with zero velocity.
Let us assume it takes time t1 to cover PQ and time t2 to cover QR.
Hence
S = UT + ½ AT2
=> 2x = 0 + ½ g(t1 + t2)^2
=> x = ¼ g(t1 + t2)^2
=> 4x = g(t1 + t2)^2 ……………..(i)
Also
x = 0 + ½ g(t1)^2
=> 2x = g(t1)^2 …………………(ii)
On dividing equation (i) by (ii),
(1)/(2) => 2 = (t1 + t2)2/(t1)2
=> [(t1 + t2)/t1]2 = 2
=> [t1+ t2/t1] = √2
=> t2/t1 = √2 – 1
=> t1 :t2 : 1 : (√2 – 1)
Tags:
Physics