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Bingel [31]
3 years ago
7

A skateboarder shoots off a ramp with a velocity of 7.3 m/s, directed at an angle of 60° above the horizontal. The end of the ra

mp is 1.5 m above the ground. Let the x axis be parallel to the ground, the +y direction be vertically upward, and take as the origin the point on the ground directly below the top of the ramp. (a) How high above the ground is the highest point that the skateboarder reaches? (b) When the skateboarder reaches the highest point, how far is this point horizontally from the end of the ramp?
Physics
1 answer:
erica [24]3 years ago
8 0

Answer:

y maximum   3.54 m, value X  2.35 m

Explanation:

We have a projectile launch problem, let's calculate the maximum height of the projectile, where the vertical speed must be zero

                     

        Vyf² = Vyo² - 2g (Y-Yo)

Where Yo is the initial height of the ramp 1.5 m

        0 = Vyo² -2g (Y-Yo)

        Y-Yo = Voy² / 2g

       Y = Yo + Voy² / 2g

Let's calculate the velocity components using trigonometry

       Voy = vo without T

       Vox = Vo cost

       Voy = 7.3 sin 60

       Vox = 7.3 cos 60

       Voy = 6.32 m / s

       Vox = 3.65 m / s

Let's calculate the maximum height

         Y = 1.5 +6.32²/2 9.8

          Y = 3.54 m

This is the maximum height from the ground

b) They ask us for the position of this point horizontally, we can calculate it looking for the time it took for the skateboarder to reach the highest point

                             

          Vfy = Voy - gt

          0 = Voy - gt

           t = Voy / g

           t = 6.32 / 9.8

           t = 0.645 s

     

Since there is no acceleration on the x-axis, we have a uniform movement, we can calculate the distance for this time

          X = Vox t

          X = 3.65 0.645

          X= 2.35 m

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A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend
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Answer:

The speed of the student just before she lands, v₂ is approximately 8.225 m/s

Explanation:

The given parameters are;

The mass of the physic student, m = 43.0 kg

The height at which the student is standing, h = 12.0 m

The radius of the wheel, r = 0.300 m

The moment of inertia of the wheel, I = 9.60 kg·m²

The initial potential energy of the female student, P.E.₁ = m·g·h₁

Where;

m = 43.0 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = 12.0 h

∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J

The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;

K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2

The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0

∴ K₁ = 0 + 0 = 0

The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0

Where;

h₂ = 0 m

The final kinetic energy, K₂, of the wheel and student is give as follows;

K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2

Where;

v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2 = m·g·h₂ + \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2

Where;

ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + \dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2

5,061.96 J = 21.5 kg × v₂² + 53.\overline 3 kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²

v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²

v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s

The speed of the student just before she lands, v₂ ≈ 8.225 m/s.

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Answer:

A) v_average = - 10 km / h,  B)    v = 1.6 m / s, v = 17.6 m / s

Explanation:

A) the average speed is the average speed of a body, if we assume that the direction of going up the hill is positive

    v₁ = 40 km / h

    v₂ = - 60 km / h

the average speed is

         v_average = \frac{v_1 + v_2}{2}

         v_average = ( 40 - 60)/2

         v_average = - 10 km / h

B) in this case they indicate the acceleration a = 3.2 m / s² and the velocity vo = 9.6 m / s

i) the speed for 2.5 s above

         v = v₀ + a t

as the time is earlier

          t = - 2.5 s

we substitute

          v = 9.6 - 3.2 2.5

          v = 1.6 m / s

ii) the velocity for a subsequent time of 2.5 s

          t = 2.5 s

           

we substitute

          v = 9.6 + 3.2 2.5

          v = 17.6 m / s

3 0
3 years ago
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