# A 200 kg roller coaster goes down its first drop at 100m. What is the coasters speed at the bottom of the hill ?

Feb 20, 2018

There are a whole host of assumptions we would have to make, but .... here goes.

#### Explanation:

OK, assuming that the change in height is the full 100m we can find the change in gravitational potential energy and (assuming no friction, air resistance, sound, vibration etc) take it that all that change in PE reappears as KE (movement) hence find the velocity.

I’m going to use the symbol $\Delta$x to mean “a change in” x.

$\Delta G P E = m \times g \times \Delta h = \Delta K E$ so

$\Delta K E = 200 \times 9.81 \times 100 = 196 , 200$J $\approx 196$ kJ

As $K E = \frac{1}{2} m \times {v}^{2}$ so $v = \sqrt{\frac{2 K E}{m}}$

So $v = \sqrt{\frac{2 \times 196200}{200}}$

Thus $v = 44.3$m/s

[Later on, you may start using the formula $v = \sqrt{2 g h}$ which does all the algebraic manipulation for you, but doesn’t show the logical steps]