To solve the problem we apply the motion kinematic equations as well as the techniques used to solve second order polynomial equations.
By definition we know that the height of a body is given by the function
[tex]h = \frac{1}{2}gt^2+v_i*t+h-0[/tex]
As there is no initial speed or a distance previously traveled, the equation is:
[tex]h = \frac{1}{2}gt^2[/tex]
Where,
g = Gravitational acceleration
t = time
h= Height
Rearranging to find the time we have,
[tex]t = \sqrt{\frac{2h}{g}}[/tex]
Then, the sound of the splash take a time [tex]\frac{h}{v}[/tex] to travel back, therefore, in time, it is necessary to adhere the new term, which converts the final time into:
[tex]t = \sqrt{\frac{2h}{g}}\frac{h}{v}[/tex]
If we make a similarity to the polynomial equation of the second degree where [tex]x = \sqrt {h}> 0[/tex] we have to:
[tex]\frac{1}{v} x^2 +\sqrt{\frac{2}{g}}x-t = 0[/tex]
Solving to find x (which is equivalent to [tex]x ^ 2[/tex]) we have to:
[tex]x = \pm 3.253[/tex]
Since the positive distance is what allows us to find the actual distance traveled we have finally to
[tex]h = x^2 = 10.58 m[/tex]
The correct answer is B.
