The concept used to solve this problem is that given in the kinematic equations of motion. From theory we know that the change in velocities of a body is equivalent to twice the distance traveled by acceleration, in other words:
[tex]v_f^2-v_i^2 = 2ax[/tex]
Where,
[tex]v_{f,i} =[/tex] Final and initial velocity
a = Acceleration
x = Displacement
For the given case, the displacement is equivalent to the height (x = h) and the acceleration is the same gravitational acceleration (a = g). In turn we do not have initial speed therefore
[tex]v_f^2 = 2hg[/tex]
[tex]v_f = \sqrt{2hg}[/tex]
Our values are given as
[tex]h = 70km = 70*10^3m[/tex]
[tex]g = 2m/s^2[/tex]
Replacing we have that,
[tex]v_f = \sqrt{2hg}[/tex]
[tex]v_f = \sqrt{2(70*10^3)(2)}[/tex]
[tex]v_f = 529.15m/s[/tex]
Therefore the speed with which the liquid sulfur left the volcano is 529.15m/s
