HGiven,
The height of the cliff, h- 20.0 m
The initial velocity of the rock, u=10.0 m/s
The final velocity of the rock, v=0 m/s
The acceleration acting on the rock is g=-9.8 m/s. The negative sign indicates that it is in the direction opposite to that of the initial velocity, i.e., downward direction.
a) From the equation of the motion, we have
[tex]v^2-u^2=2gs[/tex]Where s is the maximum height reached by the rock
On substituting the known values,
[tex]\begin{gathered} 0-10^2=2\times-9.8\times s \\ \Rightarrow s=\frac{-10^2}{2\times-9.8} \\ s=5.1\text{ m} \end{gathered}[/tex]Thus the maximum height reached by the rock above the cliff is 5.1 m
The maximum height reached by the rock from the ground is,
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