What is your speed at the bottom of the frictionless (well, almost!) Tornado water slide at Great America if the height of the slide is 32.3 meters and the length of the slide is 42.3 meters?
+6250 The easiest way to answer this is to look at the energy at the top and bottom and note that since there is no friction this value must be the same. Energy is conserved.
E = PE + KE ; potential energy + kinetic energy
PE = m g h ; mass x gravity acceleration (9.8m/s^2) x height
KE = 1/2 m v^2 ; v is velocity
Let the bottom of the slide be at height 0, and the top at height 32.3m.
At the top PE = 32.3 mg, at the bottom PE = 0.
At the top the kinetic energy is 0, at the bottom it is 1/2 m v^2
so 32.3 mg = 1/2 m v^2
32.3g = 1/2 v^2
633.08 = v^2
v=25.16 m/s
+6250 The easiest way to answer this is to look at the energy at the top and bottom and note that since there is no friction this value must be the same. Energy is conserved.
E = PE + KE ; potential energy + kinetic energy
PE = m g h ; mass x gravity acceleration (9.8m/s^2) x height
KE = 1/2 m v^2 ; v is velocity
Let the bottom of the slide be at height 0, and the top at height 32.3m.
At the top PE = 32.3 mg, at the bottom PE = 0.
At the top the kinetic energy is 0, at the bottom it is 1/2 m v^2
so 32.3 mg = 1/2 m v^2
32.3g = 1/2 v^2
633.08 = v^2
v=25.16 m/s
