Question

a 2500 kg car initially at rest rolls down a frictionless hill that is inclined at 37.2^{\circ}.

if the hill is 30.0 m long, what is the velocity of the car as it reaches the base of the hill?

a = 5.93 \text{ m/s}^2

v_f = ? \text{ m/s}

Explanation:

⚡ Using what you learned: conservation of mechanical energy

Step 1: Identify given values

$$
LATEXBLOCK0
$$

Step 2: Apply kinematic equation

$$ v_f^2 = v_i^2 + 2ad $$

$$ v_f^2 = 0^2 + 2(5.93)(30.0) $$

$$ v_f^2 = 355.8 $$

$$ v_f = \sqrt{355.8} \approx 18.9\text{ m/s} $$

Answer:

18.9