A roller coaster has a mass of 275 kg. It sits at the top of a hill with height 85 m. If it drops from this hill, how fast is it going when it reaches the bottom? (Assume there is no air resistance or friction.)

Question

shamim · Answer

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shamim · Answer

c my attachment

anonymous · Answer

Using the 3rd Equation of Rectilinear Motion, 
"V(squared) = U(squared)+ 2as " 
 intial velocity(U) is = 0, a= -g= -10 and s = -h= -85, as I take the peak as my reference. put the values in the Equation, we'll get

V(squared)=2$$\times$$ 10$$\times$$ 85 =1700 thus final velocity denoted with V = 10$$\sqrt{17}$$.

rajat97 · Answer

you can use work energy theorem
work done by gravity=change in the kinetic energy(as no other forces are working on the coaster)
therefore
mgh=kf-ki(ki = 0 as it has just started its motion)
therefore
275x85x9.8=(275xv^2)/2
thus by solving this , we get 
41.23