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# If a rollercoaster is traveling at a 310 ft hill at 20 ft/sec, what its acceleration due to gravity?

I have to find the Total Energy of my first hill of my rollercoaster using mgh+mv^2/2= Total Energy. I dont know how to find the g (the acceleration due to gravity using 32ft/sec^2) or v (the velocity of the train in ft/sec) and my teacher told me to keep m as a variable because I dont know the mass of the train. SO can someone explain to me how to find this? My hills height is 310 ft btw. Please help :(

### 1 Answer

- az_lenderLv 79 years agoFavorite Answer
The potential energy mgh is (32 ft/s^2)(310 ft)(m) when the RC is on top of the 310-ft hill.

You could also write this as (9920 ft^2/s^2) (m). That's as far as you can go with the mgh.

Does the 20 ft/s apply when the RC is on top of the hill? Or is that happening somewhere else?

Anyway, wherever the v=20 ft/s, the kinetic energy is (200 ft^2/s^2) (m).

If the 20 ft/s is happening at the top of the hill, then the total energy at the top of the hill is

(10120 ft^2/s^2) (m).