An object slides down a frictionless 33 degree incline whose vertical height is 54.0 cm. How fast is it going in meters/second when it reaches the bottom?

Answer 1

KE gained = PE lost
(1/2)mv^2 = mgh
v = sqrt(2gh)= sqrt(2*9.8*0.54)m/s

Answer 2

First of all convert 54cm to m
1m=100cm, therefore 54cm=0.54m=h. then using equation of a motion v^2=u^+2gh,since u =0 g=10 the equation bcoms v^2=2gh then substitute and solve for v ans=32.9