Answer :
Answer:
[tex]v \approx 52.421\,\frac{ft}{s}[/tex]
Explanation:
The maximum velocity can be determined by the use of the Principle of Energy Conservation:
[tex]\frac{1}{2}\cdot m_{truck}\cdot v^{2} = m_{truck}\cdot g \cdot s \cdot \sin \theta[/tex]
[tex]v = \sqrt{2\cdot g \cdot s \cdot \sin \theta}[/tex]
[tex]v = \sqrt{2\cdot (32.174\,\frac{ft}{s^{2}} )\cdot (165\,ft)\cdot \sin 15^{\textdegree}}[/tex]
[tex]v \approx 52.421\,\frac{ft}{s}[/tex]