Question

find the final speed and the time taken for a skier who skies 75 m along a slope that is 15°, measured from horizontal, in the following situations (you may neglect friction). How long (in s) does it take for the skier to reach the bottom of this hill starting from rest?

Answer 1

Answer:

V = 19.52 m/s

t = 7.69 seconds

Explanation:

Given:-

- The total distance of ski track, L = 75 m

- The descent of the ski track, θ = 15°

Find:-

find the final speed and the time taken

Solution:-

- Assuming the fictitious forces (Drag, Friction) acting on the skier are negligible, then the skier can be considered as a system on which there are no unbalanced forces.

- Taking the mass of the skier = m

- Consider the initial point where, velocity Vi = 0.

- Then apply the principle of conservation of energy, where all the potential energy at the initial point is converted into kinetic energy.

                              ΔK.E = ΔP.E

                              Kf - Ki = Pf - Pi

Where,  Kf = final kinetic energy = 0.5*m*V^2

             Ki = Initial kinetic energy = 0

             Pf = Final potential energy = 0 ... Datum

             Pi = Initial potential energy =  - m*g*sin ( θ )*L

                            0.5*m*V^2 - 0 = 0 - (- m*g*sin ( θ ))*L

                            V = √2*g*L*sin ( θ )

                            V = √2*9.81*75*sin ( 15 )

                            Vf = 19.52 m/s

- Now use the first kinematic equation of motion for the skier skiing down the slope:

                          Vf = Vi + a*t

Where,   a = Acceleration down the slope = g*sin ( θ )

               t = Time taken

                          Vf = g*sin ( θ )*t

                          t = 19.52 / [ 9.81*sin ( 15 ) ]

                          t = 7.69 seconds

Answer 2

Answer:

Final speed = 148.21m.s

Time of flight = 7.82seconds

Explanation:

The motion of the body is a projectile motion.

Projectile is a motion created by an object launched in air and allowed to fall to freely under the influence of gravity.

Taking the maximums height reached H = 75m

Angle of launch = 15°

Using the maximum height formula to get the velocity U of the object

H = U²sin²theta/2g

Where g is the acceleration due to gravity = 9.81m/s²

75 = U²(sin15°)²/2(9.81)

1471.5 = U²(sin15°)²

1471.5 = 0.06699U²

U² = 1471.5/0.06699

U² = 21,965.9

U = √21,965.9

U = 148.21m/s

The time taken for the skier to reach the bottom of this hill starting from rest is the time of flight T.

T = 2Usintheta/g

T = 2(148.21)sin15°/9.81

T = 296.42sin15°/9.81

T = 76.72/9.81

T = 7.82seconds