A stone of mass 4gm is …

For the 1st part the stone is falling under gravity. Let t₁ be the time from release to splashdown.

d =depth

So d= gt₁²/2  -  (1)

After splashdown the sound travels back up. Using 330m/s for the speed of sound this gives:

d=330×t₂  (2)

We also know that:

t₁+t₂=4s  (3)

Combining (1)and(2) we get:

330t₂= gt₁²/2

From (3)⇒

t₂=(4−t₁)

So:

330(4−t₁)=gt₁²/2

5t₁²+330t-1320=0

t₁²+66t-264=0

Solving this quadratic equation using Discriminant formula and ignoring the -ve root we get

t₁=3.78s

So t2=4−3.78=0.22s

So d=330×t₂=330×0.22=72.6m

this is $\frac{19h}{85} instead of \frac{19h}{5}$

Ans. Let Depth of Well = h 

$t_{stone} + t_{sound} = 4 .................. \left(1\right)$

$For t_{stone }(free-falling body ) :$

$h = \frac{1}{2}g{t}^2=> t = \sqrt{\frac{2h}{g}}$

$t_{stone}= \sqrt{\frac{2h}{10}}$

= > $t_{stone }= \sqrt{\frac{h}{5}}$

$For t_{sound }(uniform motion):$

$h = ut=> t = \frac{h}{u}$

$t_{sound} = \frac{h}{340}$

Put Both values  values in eq (1), We get 

$\sqrt{\frac{h}{5}} + \frac{h}{340} = 4$

=> $\sqrt{\frac{h}{5}} = 4 - \frac{h}{340}$

Squaring both sides

$\frac{h}{5} = 16 + \frac{h^2}{{340}^2} - \frac{2h}{85}$

=> $\frac{h^2}{115600} + 16 = \frac{h}{5} + \frac{2h}{85}$

=> $\frac{h^2}{115600} - \frac{19h}{5} + 16 = 0$

Solve for h and you ll have answer