Respuesta :
Given data
*The given initial elevation is h = 62.5 m
*The lowest point above the ground is H = 5.0 m
The formula for the final speed of the roller coaster is given by the conservation of energy as
[tex]\begin{gathered} U_i=U_f \\ mgh+\frac{1}{2}mv_i^2=mgH+\frac{1}{2}mv^2_f^{} \end{gathered}[/tex]*Here U_i is the initial energy
*Here U_f is the final energy
*Here v_i = 0 m/s is the initial speed
*Here g is the acceleration due to the gravity
Substitute the known values in the above expression as
[tex]\begin{gathered} mg(62.5)+\frac{1}{2}m(0)^2=mg(5.0)+\frac{1}{2}mv^2_f \\ v=\sqrt[]{115g} \\ =33.57\text{ m/s} \end{gathered}[/tex]As from the given data, the vertical acceleration at the bottom point does not exceed (2g) as
[tex]\begin{gathered} a\leq2g \\ \frac{v_f^2}{r}\leq2g \\ r\ge\frac{v_f^2}{2g} \\ r\ge\frac{(33.57)^2}{2(9.8)} \\ r\ge57.5\text{ m} \end{gathered}[/tex]Hence, the radius of the loop is 57.5 m
