Question

a 200 kg roller coaster car is located at the top of a hill where the height is 30 m. what is the velocity when it reaches the bottom of the hill?
5.5 m/s
11.2 m/s
16.7 m/s
24.2 m/s

Explanation:

Step1: Apply conservation of energy

$PE_{top}=KE_{bottom}$

Step2: Write potential - energy formula

$PE = mgh$, where $m = 200$ kg, $g=9.8$ m/s², $h = 30$ m

Step3: Write kinetic - energy formula

$KE=\frac{1}{2}mv^{2}$

Step4: Equate and solve for $v$

$mgh=\frac{1}{2}mv^{2}$, cancel out $m$ on both sides. We get $gh=\frac{1}{2}v^{2}$. Then $v=\sqrt{2gh}$. Substitute $g = 9.8$ m/s² and $h = 30$ m into the formula: $v=\sqrt{2\times9.8\times30}=\sqrt{588}\approx24.2$ m/s.

Answer:

24.2 m/s