Respuesta :
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
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
