# Physics

A roller-coaster car is moving at 20.0 m/s along a straight horizontal track. What is its speed after climbing a 15.0 m hill? Neglect the effects of friction.

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1. Without friction and air resistance, we equate potential (PE) and kinetic (KE)energies.

Energies at bottom of hill = energies at top of hill.

0+(1/2)m*vi²=mgh+(1/2)m*vf^sup2;

cancel m and substitute
g=9.8 m/s²
vi=20 m/s
h=15 m

(1/2)(20²)=9.8*15+(1/2)vf²

Solve for vf.

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2. HELP !! I do not fully understand how to completely solve this problem. Can you please help me further with the equation?? Thank You. !

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