Question

if the skater starts at a height of 2h how will her velocity at the bottom of the ramp compare?a vb 1.4vc 2vd 4v

Explanation:

Step1: Set energy conservation (start)

Gravitational potential energy: $PE_1 = mg(2H)$
Kinetic energy: $KE_1 = 0$ (starts at rest)
Total energy: $E_1 = 2mgH$

Step2: Set energy conservation (bottom)

Gravitational potential energy: $PE_2 = 0$ (reference level)
Kinetic energy: $KE_2 = \frac{1}{2}mv^2$
Total energy: $E_2 = \frac{1}{2}mv^2$

Step3: Equate total energies

By energy conservation: $E_1 = E_2$
$2mgH = \frac{1}{2}mv^2$

Step4: Solve for velocity v

Cancel $m$, rearrange: $v^2 = 4gH$
$v = 2\sqrt{gH}$
(Assume velocity at height $H$ is $V = \sqrt{gH}$, so $v=2V$)

Answer:

C. 2V