Astudent is skateboarding down a ramp that is 8.51 m long and inclined at 23.7° with respect to the horizontal. the initial speed of the skateboarder at the top of the ramp is 4.97 m/s. neglect friction and find the speed at the bottom of the ramp.

vf = 9.58 m/s

Explanation:

let vf be the velocity of the students at the bottom of the ramp, Kf be the kinetic energy of the student at the bottom of the ramp and vi be the velocity of the student at the top of the ramp as well as Ki be the kinetic energy of the student at the top of the ramp. let m be the mass of the skateboarder and Wf be the work done by gravity and Wtot be the total work done on the skateboarder, let Ф be the inclined angle of the ramp and α be the angle between the force and the path of the motion. and ΔX be the distance from the top of the ramp to the bottom that the skateboarder will cover.

from Work and Energy principles:

Wtot = Kf - Ki

if we ignore friction, the only force doing work is the force of gravity(parallel component), then:

Wf = Kf - Ki

Fg×sin(Ф)×ΔX×cos(α) = 1/2(m)(vf)^2 - 1/2(m)(vi)^2

because the force is in the same direction as the path of the motion,   cos(α) = 1;

Fg×sin(Ф)×ΔX = 1/2(m)(vf)^2 - 1/2(m)(vi)^2

m×g×sin(Ф)×ΔX = 1/2(m)(vf)^2 - 1/2(m)(vi)^2

    g×sin(Ф)×ΔX = 1/2(vf)^2 - 1/2(vi)^2

2×g×sin(Ф)×ΔX =  (vf)^2 - (vi)^2

(vf)^2  = 2×g×sin(Ф)×ΔX + (vi)^2

          = 2×(9.8)×sin(23.7)×(8.51) + (4.97)^2

          = 91.744

vf =  \sqrt{91.744}  = 9.58 m/s

Therefore, the speed of the scateboarder at the bottom of the ramp is 9.58 m/s.

dang that balls went far