Answer:
The cliff is 12.632 meter high.
Explanation:
Each stone experiments a free fall motion, that is, an uniformly accelerated motion due to gravity. We construct the respective equations of motion for each stone:
First stone
[tex]y_{1} = y_{o,1} +v_{o,1}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}[/tex] (1)
Second stone
[tex]y_{2} = y_{o,2} +v_{o,2}\cdot (t-2) +\frac{1}{2}\cdot g\cdot (t-2)^{2}[/tex] (2)
Where:
[tex]y_{1}[/tex], [tex]y_{2}[/tex] - Final height of the first and second stone, in meters.
[tex]y_{o,1}[/tex], [tex]y_{o,2}[/tex] - Initial height of the first and second stone, in meters.
[tex]v_{o,1}[/tex], [tex]v_{o,2}[/tex] - Initial speed of the first and second stone, in meters per second.
[tex]t[/tex] - Time, in seconds.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
If we know that [tex]y_{o,1} = y_{o,2}[/tex], [tex]y_{1} = y_{2} = 0\,m[/tex], [tex]v_{o,1} = 0\,\frac{m}{s}[/tex], [tex]v_{o,2} = -30\,\frac{m}{s}[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], then we find that time when both stones hit the ground simultaneously is:
[tex]4.904\cdot t^{2} = -30\cdot (t-2)+4.904\cdot (t-2)^{2}[/tex]
[tex]4.904\cdot t^{2} = -30\cdot t + 60 +4.904\cdot (t^{2}-4\cdot t +4)[/tex]
[tex]-30\cdot t +60 -19.616\cdot t +19.616 = 0[/tex]
[tex]49.616\cdot t = 79.616[/tex]
[tex]t = 1.605\,s[/tex]
The height of the cliff is:
[tex]y_{1} = y_{o,1} +v_{o,1}\cdot t +\frac{1}{2}\cdot g\cdot t^{2}[/tex]
[tex]y_{o,1} = 12.632\,m[/tex]
The cliff is 12.632 meter high.
