Here , we can directly use the formula for distance covered in nth second
i.e., sth= u+a/s(2n-1)
therefore...... For 2nd second ,
S2nd=u+10/2(2×2-1) =u+ 15 …...........(1)
similarly...........For 9th second
S9th=0+10/2(2×9-1) = 65 …............(2)
( in the later case i have put u=0 because body is returning back after reaching to its highest point ,and at highest point velocity is zero)
Now equating 1 and 2 we will get
u= 50m/s
now usinng equation third of newton’s law of motion
v2= u2 + 2as
(0)2 = (50)2 + 2 × 10 × s
s= 125 m
