Answer to Question #95299 in Mechanics | Relativity for Soo

Question

A Hot Wheels car is rolling along a horizontal track at speed vo = 6.0 m/s. It then comes to a ramp inclined at an angle 0 = 30° above the horizontal, and the car undergoes a deceleration of g sin 0 = 4.9 m/s2 when moving along the ramp. The track ends at the top of the ramp, so the car is launched into the air. By the time the car reaches the top of the ramp, its speed has gone down to 3.0 m/s. (a) How high is the top of the ramp (vertical height, not distance along the ramp)?

Answer 1

The law of conservation of energy gives

$$\frac{mv_i^2}{2}=\frac{mv_f^2}{2}+mgh$$

So, the height of the ramp

$$h=\frac{v_i^2-v_f^2}{2g}=\frac{6.0^2-3.0^2}{2\times 9.8}=1.38\:\rm m$$