A stone is dropped into a well of 20 m deep. Another stone is thrown downward with velocity ‘v’ one second later. If both stones reach the water surface in the well simultaneously, v is equal to (g = 10 ms^-2)

First stone was per forming a free fall 

so time taken by it to cover the depth h = 20m will be

t= $\sqrt{2h/g}$​ just by using h = 1/2​gt²​

t = $\sqrt{2\times 20/10}$ ​​= 2 second

Second stone thrown 1 second later 

reaches simultaneously with first stone

so time taken will be t−1=2−1=1 second

the second stone will follow

h = vt_1​ + 1/2​gt₁²​

putting h = 20m, g = 10m/s² & t_1​ = 1sec

we get 

20 = v + 5×1

⇒v = 15m/s​