Brittany is changing the tire of her car on a steep hill 22.4 m high. She stumbles and drops the 10.5-kg spare tire, which rolls down the hill, starting with an initial speed of 2.10 m/s.
What is the speed of the tire at the top of the next hill, which is 8.59 m high?

If we assume that there is no friction between the tire and the ground, we can the use formula:
ΔKE + ΔPE = 0
where KE is the kinetic energy which is equal to 0.5mv²
and PE is the potential energy which is equal to mgh

Substituting the given values:
0.5 (10.5) (v2 - 2.10^2) + 10.5 (9.81) (8.59 - 22.4) = 0

Solving for v:
v = 16.59 m/s

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Brittany is changing the tire of her car on a steep hill 22.4 m high. She stumbles and drops the 10.5-kg spare tire, which rolls down the hill, starting with an initial speed of 2.10 m/s. What is the speed of the tire at the top of the next hill, which is 8.59 m high?

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Brittany is changing the tire of her car on a steep hill 22.4 m high. She stumbles and drops the 10.5-kg spare tire, which rolls down the hill, starting with an initial speed of 2.10 m/s.
What is the speed of the tire at the top of the next hill, which is 8.59 m high?