A stone is dropped from height H and after 1 second, another stone id dropped from a balcony 20 m below.They reach the ground simultaneously?

Let H be the height. Time taken is such that

H= 1/ g t^2 (1)

Now we have the second equation

H-20= 1/ g (t-1)^2

H-20 = 1/2 g (t^2-2t+1) (2)

Subtract (1) from (2), you get

-20 = 1/2 g ( 1- 2t)

Rearrange the term to get

t=49.8/19.6 =2.54.

Height of the building is

H = 1/ g (2.54)^2= 31.6 m.

time for 2nd stone to reach ground = t = √2h/g = √[2(20)/9.8] = 2.02 s

time for 1st stone to reach ground = t + 1 = 2.02 + 1 = 3.02 s

H = 1/2g(t+1)² = (0.5)(9.8)(3.02)² = 44.7 ≈ 45 m ANS