If the lift leaves the base at a velocity of 15.5 m/s and arrives at the top with a final velocity of 0 m/s, what is the height of the hill?

Question

Round to the nearest tenth

Answer 1

Equate total energy at the base to total energy at the top, and solve for height.

 

mv₁²/2 + mgh₁ = mv₂²/2 + mgh₂

 

h₁ = 0  [base]

v₂ = 0  [come to rest at top]

 

v₁²/2 = gh₂

 

h₂ = v₁²/(2g) = (15.5)²/[2(9.8)] meters = ? meters