The Raging Bull approaches the bottom of a hill on its way up at a speed of 35.9 m/s. The hill is 17.9 m high, and the riders are screaming all the way to the top! Ignoring air resistance and friction, find the speed at which the riders go flying over the top of the hill!
+33654 This requires conservation of energy. Kinetic energy at the bottom of the hill equals the sum of kinetic and potential energies at the top.
KE at bottom = (1/2)mv12, where m is mass and v1 is velocity at bottom (35.9ms-1)
PE at top = mgh, where g is gravitational acceleration (9.81ms-2) and h is height if hill (17.9m)
KE at top = (1/2)mv22, so
(1/2)mv22 + mgh = (1/2)mv12
The mass cancels throughout, so
(1/2)v22 +9.81*17.9 = (1/2)35.92,
Multiply throughout by 2
v22 + 2*9.81*17.9 = 35.92,
See if you can take it from here!
+33654 This requires conservation of energy. Kinetic energy at the bottom of the hill equals the sum of kinetic and potential energies at the top.
KE at bottom = (1/2)mv12, where m is mass and v1 is velocity at bottom (35.9ms-1)
PE at top = mgh, where g is gravitational acceleration (9.81ms-2) and h is height if hill (17.9m)
KE at top = (1/2)mv22, so
(1/2)mv22 + mgh = (1/2)mv12
The mass cancels throughout, so
(1/2)v22 +9.81*17.9 = (1/2)35.92,
Multiply throughout by 2
v22 + 2*9.81*17.9 = 35.92,
See if you can take it from here!
