if a particle takes t seconds less and acquires a velocity of vm/s more in falling through the same distance on two planets whe accelarations due to gravity are 2g and 8g respectively the what is the value of v?

let u1(t) and u2(t) be the speeds on 2g and 8g planets respectively

let motion start at t0,fall start at T,and T1(8g), and falls end at t1, and t2(8g)

let fall length be d,

u1(T)=2gT, u2(T)=8gT,u1(t1)=2g(t1),and u2(t2)=8g(t2)

d=(u1(T)+u1(t1))/(2(t1-T)

d=(u2(T1)+u2(t2))/(2(t2-T1)

t2-T1=t

u1(T1)-u2(t2)=v

the answer is 4gt. can anyone explain how? 

thank you in advance.

U v to take only 3 equations in consideration. Here r d following

Assume v₀ be the velocity achieved by the particle in 2g . In 8g it will achive a velocity of v₀+v 

Assume that the particle take t₀ time in 8g. Then it had took a time of t+t₀ in 8g.

Using Newtons first eq.

 v₀ = 2g(t+  t₀) _---------(I)

(v+ v₀)= 8gt -------(II)

Solving the pair of equation gives v=6g(t-t₀)  ---(III)

Using Newtons second equation .

Assume the height through which they fall = H

H =1/2(2g)( t+t₀)^2 and H= 1/2(8g) (t₀)^2 ==> 1/2(2g)( t+t₀)^2 = 1/2(8g) (t₀)^2

 ==> (t₀)= t/3 

Substituting value of  (t₀) in eq III gives

v= 6g(2t/3) = 4gt