Answer: -17.66 m/s or 17.66 m/s downwards
Explanation:
The decribed situation is related to vertical motion and the main equation is:
[tex]y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}[/tex] (1)
Where:
[tex]y=0 m[/tex] is the final height of both stones
[tex]y_{o}=75 m[/tex] is the initial height of both stone s
[tex]V_{o}[/tex] is the initial velocity of the stone
In the case of the first stone is [tex]V_{o1}=0 m/s[/tex] and for the second stone [tex]V_{o2}[/tex] is different from zero
[tex]t[/tex] is the time, which is the same for both stones
[tex]g=-9.8 m/s^{2}[/tex] is the acceleration due to gravity
Having this clear, let's find [tex]t[/tex] for the first stone:
[tex]y=y_{o}+V_{o1}t-\frac{1}{2}gt^{2}[/tex] (2)
[tex]0=y_{o}+0-\frac{1}{2}gt^{2}[/tex] (3)
Isolating [tex]t[/tex]:
[tex]t=\sqrt{\frac{-2y_{o}}{g}}[/tex] (4)
[tex]t=\sqrt{\frac{-2(75 m)}{-9.8 m/s^{2}}}[/tex] (5)
[tex]t=3.912 s[/tex] (6)
Now we have to use this time to find [tex]V_{o2}[/tex]:
[tex]0=y_{o}+V_{o2}t-\frac{1}{2}gt^{2}[/tex] (7)
Isolating [tex]V_{o2}[/tex]:
[tex]V_{o2}=-\frac{y_{o}}{t} - \frac{gt}{2}[/tex] (8)
[tex]V_{o2}=-\frac{75 m}{3.912 s} - \frac{(-9.8 m/s^{2})(3.912)}{2}[/tex] (9)
Finally:
[tex]V_{o2}=-17.633 m/s[/tex] (10) The negative sign indicates the direction of the initial velocity is vertically down
