A stone thrown vertically up with velocity v reaches three points A, B and C with velocities $\mathrm{v},\frac{\mathrm{v}}{2}$ and $\frac{\mathrm{v}}{4}$ respectively. Then AB:BC is

v² = u²−2gs

(v/2​)² = v²−2gs₁​

s₁ ​= 3​v²/4(1/2g​) at point A

(v/4​)² = v²−2gs_2​

s₂ ​= 15v²​/16(1/2g​)  at point B

(v/8​)² = v² − 2gs₃​

s₃​ = 63/64v²​(1​/2g)  at point C

AB:BC = (s₂​−s₁​):(s₃​−s₂​)

⇒$\frac{\left(\frac{v^2}{2g}\right)\left(\frac{15}{16}-\frac{3}{4}\right)}{\left(\frac{v^2}{2g}\right)\left(\frac{63}{64}-\frac{15}{16}\right)}$

⇒4×3/3 ​= 4:1