At the top, weight = centripetal force
mg=mv^2/r
solve for v
A stuntman drives a car over the top of a hill, the cross section of which can be approximated
by a circular arc of radius 250 m. What is the greatest speed at which he can drive without
the car leaving the road at the top of the hill?
3 answers
The normal force and the centripetal *force need to equal eachother if the car is to remain grounded, so N=mV^2/r or mg=mV^2/r and then the masses cancel and we have g=V^2/r or v^2=gr.
49.5 m/s
4 answers