Answer:
a) The initial speed of the rock is approximately 14.607 meters per second.
b) The greatest height of the rock from the base of the cliff is 42.878 meters.
Explanation:
a) The rock experiments a free-fall motion, that is a vertical uniform accelerated motion due to gravity, in which air friction and effects of Earth's rotation. By Principle of Energy Conservation we have the following model:
[tex]U_{g,1}+K_{1} = U_{g,2}+K_{2}[/tex] (Eq. 1)
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energies, measured in joules.
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energies, measured in joules.
By definitions of gravitational potential and translational kinetic energies we expand and simplify the equation above:
[tex]m\cdot g\cdot (y_{1}-y_{2})= \frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2})[/tex]
[tex]g\cdot (y_{1}-y_{2}) = \frac{1}{2}\cdot (v_{2}^{2}-v_{1}^{2})[/tex] (Eq. 2)
Where:
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y_{1}[/tex], [tex]y_{2}[/tex] - Initial and final height, measured in meters.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final speed of the rock, measured in meters per second.
If we know that [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 32\,m[/tex], [tex]y_{2} = 0\,m[/tex] and [tex]v_{2} = 29\,\frac{m}{s}[/tex], then the equation is:
[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (32\,m-0\,m) = \frac{1}{2}\cdot \left[\left(29\,\frac{m}{s} \right)^{2}-v_{1}^{2}\right][/tex]
[tex]313.824 = 420.5-0.5\cdot v_{1}^{2}[/tex]
[tex]0.5\cdot v_{1}^{2} = 106.676[/tex]
[tex]v_{1} \approx 14.607\,\frac{m}{s}[/tex]
The initial speed of the rock is approximately 14.607 meters per second.
b) We use (Eq. 1) once again and if we know that [tex]g =9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{1} = 32\,m[/tex], [tex]v_{1} \approx 14.607\,\frac{m}{s}[/tex] and [tex]v_{2} = 0\,\frac{m}{s}[/tex], then the equation is:
[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (32\,m-y_{2}) = \frac{1}{2}\cdot \left[\left(0\,\frac{m}{s} \right)^{2}-\left(14.607\,\frac{m}{s} \right)^{2}\right][/tex]
[tex]313.824-9.807\cdot y_{2} = -106.682[/tex]
[tex]9.807\cdot y_{2} = 420.506[/tex]
[tex]y_{2} = 42.878\,m[/tex]
The greatest height of the rock from the base of the cliff is 42.878 meters.
