Albert Jr. takes his new skateboard out to the top of the loading ramp. The ramp is 30 m long and sloped at an angle of 15 degrees relative to the horizontal. Assuming he starts from rest, and neglecting friction, about how fast is Albert Jr. going at the bottom of the ramp?

vf = 12.343 m/s

Explanation:

Given:

- The Length of the ramp s = 30 m

- The angle of the ramp θ = 15 degrees

- Initial velocity at the top of ramp vi = 0

- Neglect Friction

Find:

How fast is Albert Jr. going at the bottom of the ramp?

Solution:

- We can use conservation of energy for the skateboard at top and bottom of the ramp.

                                ΔK.E = ΔP.E

Where, ΔK.E : Change in kinetic energy

            ΔP.E : Change in gravitational potential energy

                                0.5*m*(vf^2 - vi^2) = m*g*Δh

Where, m: The mass of the object

            Δh: Change in elevation from top to bottom

                                Δh = s*sin(θ)

- Substitute and simplify:

                                 0.5*(vf^2 - vi^2) = g*s*sin(θ)

                                 vf^2 = 2*g*s*sin(θ)

                                 vf = √(2*g*s*sin(θ)) = √(2*9.81*30*sin(15))

                                 vf = 12.343 m/s

- The velocity at the bottom of ramp would be 12.343 m/s

the right answer is 0.3 meters per second

the ima (higher than 1m) will be greater than the ima with the height of 1m.