Respuesta :
Answer:
v = sqrt[2*(F*h*cot(theta)-mgh)/m]
Explanation:
Work = KE + Ug
F*r=1/2mv^2+mgh
1/2mv^2=F*r-mgh
v=sqrt[2(F*r-mgh)/m]
r=h/tan(theta)=h*cot(theta)
Answer:
The correct answer is [tex]\sqrt{[2*(\frac{F*h*cot(Q)-mgh)}{m} ]}[/tex]
Explanation:
The work done is defined as the sum of the kinetic energy and the potential energy.
Work = KE + Mgh
[tex]F*r=\frac{1}{2} mv^2+mgh[/tex]
[tex]\frac{1}{2} mv^2=F*r-mgh[/tex]
[tex]v=\sqrt{[\frac{2(F*r-mgh)/}{m} ]}[/tex]
[tex]r=\frac{h}{tanQ} \\r=h*cot(Q)[/tex]
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https://brainly.com/question/19500270?referrer=searchResults
