the distance travelled by a body during last second of its total flight is 'd'when the body is projected vertically up with certain velocity . If the velocity of projection is doubled, the distance travelled by the body during last second of its total flight is

 Let the total time of flight in first case in t.....

 After reaching the highest point particle will have velocity equal to zero.....and will take time t/2 to come to the ground....

 now t=2u/g   

 so time to come down from peak = u/g

 

Distance travelled in t=u/g will be  u²/2g...

 and in (t-1 )  it will be  g(t-1)²/2  = (u-g)²/2g..

 so d = u²/2g - (u-g)²/2g

 d = (2u-g)/2 --------------(1)

and if projection velocity is doubled.....then we get D= (4u-g)/2

 

from (1)

2d = 2u-g

 

4u = 2(2d+g)

 

4u - g = 4d + g 

surely  D = 2d + g/2