A 60 kg cyclist approaches the bottom of a gradual hill at a speed of 11 m/s. The hill is 5.0 m high, and the cyclist estimates that she is going fast enough to coast up and over it without peddling. Using conservation of energy, find the speed of the cyclist when she reaches the top of the hill (ignoring air resistance and friction). g

9.91m/s

Explanation:

According to conservation of energy, the potential energy at the top of the hill is equal to the kinetic energy along the hill.

PE = KE

mgh = 1/2mv²

m is the mass of the cyclist

g is the acceleration due to gravity

h is the height of the hill

v is the speed of the cyclist

From the formula:

gh = v²/2

Given

g = 9.81m/s²

h = 5.0m

Substitute into the formula:

9.81(5.0) =  v²/2

9.81*5*2 =  v²

98.1 =  v²

v = √98.1

v = 9.91m/s

Hence the speed of the cyclist when she reaches the top of the hill is 9.91m/s

your answer is c! hope this : )

conduction, convection, radiation