ICSE Class 10 Answered

 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