When a 236 kg roller coaster car is released from rest height of 40 meters what the veloity when it gets to 18 meters?

The change in height of the roller coaster car is:

$$\Delta y = y_f - y_i = -22 \text{ m}$$

The change in potential energy of the roller coaster car is:

$$\Delta U_g = mgy_f - mgy_i = -(236 \text{ kg})(9.8 \text{ m/s}^2)(-22 \text{ m}) = 48992 \text{ J}$$

The change in kinetic energy of the roller coaster car is:

$$\Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \frac{1}{2}(236 \text{ kg})v_f^2 - 0$$

By the conservation of energy, we have:

$$\Delta U_g = \Delta K$$

$$-(236 \text{ kg})(9.8 \text{ m/s}^2)(-22 \text{ m}) = \frac{1}{2}(236 \text{ kg})v_f^2$$

Solving for $v_f$, we get:

$$v_f = \sqrt{\frac{2(48992 \text{ J})}{236 \text{ kg}}} = 28.2 \text{ m/s}$$

Therefore, the velocity of the roller coaster car when it gets to 18 meters is 28.2 m/s.