Answer :
The maximum speed of the cycle is 21 m/s.
When the motorcyclist takes the curve on the hill, a net frictional force acts on it towards the center and acts as the centripetal force.
The radius of the hill is = 45 m
Also , the weight of the motorcycle acts downwards and the normal reaction acts upwards. As the motorcycle is on the curve so , a component of the weight acts towards the center.
weight is given by , W = mg
So the net F(friction) = mg - N = [tex]\frac{mv^{2} }{r}[/tex]
For loosing contact, N=0
[tex]v^{2}[/tex] = rg
v = [tex]\sqrt{rg}[/tex]
v = [tex]\sqrt{(45)(9.80)}[/tex]
v = 21 m/s
So the maximum speed of the motorcycle can be 21 m/s.
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