As per the given question,
Height of the inclined plane "(h)=500ft"
Coefficient of friction "(\\mu)=0.02"
Gradient "(m)=\\tan\\theta=2.5"
"\\Rightarrow \\frac{height}{base}=2.5"
"\\Rightarrow \\frac{500}{base}=2.5"
"\\Rightarrow base=\\dfrac{500}{2.5}ft"
"\\Rightarrow base=\\frac{152.4}{2.5}m"
"\\Rightarrow base=60.96m"
"\\sin \\theta=\\frac{152.4}{164.13}=0.928"
"\\cos\\theta =\\frac{60.96}{164.13}=0.37"
Let the resultant acceleration of the truck is "=a"
So, "a=g\\sin\\theta-\\mu g\\cos\\theta"
"\\Rightarrow a=9.8\\times 0.92-0.02\\times 9.8\\times 0.37 m\/sec^2"
"\\Rightarrow a=8.943 m\/sec^2"
"\\Rightarrow v^2=u^2+2as"
"\\Rightarrow v=\\sqrt{2as}"
"\\Rightarrow v=\\sqrt{2\\times 8.94\\times 164.13}"
"\\Rightarrow v=54.17m\/sec"
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