Respuesta :
Answer:
1. The force of friction [tex]F_f =0.72 \times F_n[/tex]
2. The total travel time is approximately 0.884 seconds
The range is approximately 2.209 m
The maximum height reached is approximately 0.383 m
3. The time it will take the cyclist to stop is approximately 25.806 seconds
4. The time it will take the ball to reach the ground is approximately 2.020 seconds
Explanation:
1. Given that the coefficient of static friction, μ = 0.72, we have;
[tex]F_f = \mu \cdot F_n = 0.72 \times F_n[/tex]
2. The given parameters are;
The initial velocity of the car, v = 5 m/s
The direction of the vehicle = 60° to the horizontal
Therefore, we have;
The vertical component of the velocity, [tex]v_y[/tex] = 5 × sin(60°) = 2.5×√3
[tex]v_y[/tex] = 2.5×√3 m/s
The total travel time, t = 2·u·sin(θ)/g = 2 × 5 × sin(60°)/9.8 ≈ 0.884
The total travel time, t ≈ 0.884 seconds
The range, R = (u²·sin(2·θ))/g = 5² × sin(120°)/9.8 ≈ 2.209
The maximum height is given from the v² = [tex]u_y[/tex]² - 2·g·h
Where;
v = The final velocity = 0 m/s at maximum height
h = [tex]h_{(max)}[/tex] = The maximum height
[tex]u_y[/tex] = The initial vertical velocity = 2.5·√3
∴ 0² = 2.5·√3² - 2 × 9.8 × [tex]h_{(max)}[/tex]
2.5·√3² = 2 × 9.8 × [tex]h_{(max)}[/tex]
[tex]h_{(max)}[/tex] = 2.5·√3²/(2 × 9.8) ≈ 0.383
The maximum height = [tex]h_{(max)}[/tex] = 0.383 m
3. The given parameters are;
The initial velocity of the cyclist, v₀ = 16 m/s
The deceleration of the cyclist, a = -0.62 m/s²
At the point the cyclist stops, the final velocity, [tex]v_f[/tex] = 0
Therefore, from the given kinematic equations, we have;
[tex]v_f[/tex] = v₀ + a·t
Where t, is the time it will take the cyclist to stop
∴ t = ([tex]v_f[/tex] - v₀)/a = (0 - 16)/(-0.62) ≈ 25.806
Therefore, the time it will take the cyclist to stop = t ≈ 25.806 seconds
4. The time, t, it will take the ball to reach the ground, from a height, h = 20 meters is given by the equation of free fall as follows;
h = 1/2·g·t²
Where;
g = The acceleration due to gravity = 9.8 m/s²
Therefore, we have;
t² = h/(1/2·g)
t = √(h/(1/2·g)) = √(20/(1/2 × 9.8)) = 2.0203050891 ≈ 2.02
The time it will take the ball to reach the ground, t ≈ 2.020 seconds.
