Answer:
a) The vertical component of the kangaroo's velocity at take off is approximately 4.982 meters per second.
b) The angle at which the kangaroo took off the ground is approximately 46.066º.
Explanation:
a) According to the statement, the kangaroo has an initial horizontal velocity and jumps to overcome the fence until velocity becomes zero. If effects from non-conservative forces can be neglected, then we represent the situation of the kangaroo by Principle of Energy Conservation:
[tex]K_{1}+U_{g,1} = K_{2}+U_{g,2}[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final translational kinetic energies, measured in joules.
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energies, measured in joules.
By definitions of translational kinetic and potential gravitational energies, we expand and simplify the equation above:
[tex]\frac{1}{2}\cdot m\cdot [(v_{1,x}^{2}+v_{1,y}^{2})-(v_{2,x}^{2}+v_{2,y}^{2})] = m\cdot g \cdot (y_{2}-y_{1})[/tex]
[tex]\frac{1}{2}\cdot [(v_{1,x}^{2}+v_{1,y}^{2})-(v_{2,x}^{2}+v_{2,y}^{2})] = g \cdot (y_{2}-y_{1})[/tex] (Eq. 2)
Where:
[tex]m[/tex] - Mass of the kangaroo, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]y_{1}[/tex], [tex]y_{2}[/tex] - Initial and final heights of the kangaroo above the ground, measured in meters.
[tex]v_{1,x}[/tex], [tex]v_{1,y}[/tex] - Vertical and horizontal initial velocities, measured in meters per second.
[tex]v_{2,x}[/tex], [tex]v_{2,y}[/tex] - Vertical and horizontal final velocities, measured in meters per second.
If we know that [tex]v_{1,x} = 4.80\,\frac{m}{s}[/tex], [tex]v_{2, x} = 0\,\frac{m}{s}[/tex], [tex]v_{2,y} = 0\,\frac{m}{s}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]y_{2}-y_{1} = 2.44\,m[/tex], then (Eq. 2) is reduced into this:
[tex]11.52+0.5\cdot v_{1,y}^{2}=23.929[/tex]
Lastly, we solve for [tex]v_{1,y}[/tex]:
[tex]0.5\cdot v_{1,y}^{2}=12.409[/tex]
[tex]v_{1,y}^{2} = 24.818[/tex]
[tex]v_{1,y} \approx 4.982\,\frac{m}{s}[/tex]
The vertical component of the kangaroo's velocity at take off is approximately 4.982 meters per second.
b) The angle at which the kangaroo took off the ground ([tex]\theta[/tex]), measured in sexagesimal degrees, is obtained by the following inverse trigonometric relation:
[tex]\theta =\tan^{-1}\left(\frac{v_{1,y}}{v_{1,x}} \right)[/tex] (Eq. 3)
[tex]\theta = \tan^{-1}\left(\frac{4.982\,\frac{m}{s} }{4.80\,\frac{m}{s} } \right)[/tex]
[tex]\theta \approx 46.066^{\circ}[/tex]
The angle at which the kangaroo took off the ground is approximately 46.066º.
