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At t = 0, a rock is projected from ground level with a speed of 15.0 m/s and at an angle of 53.0° above the horizontal. Neglecting air resistance, what is the initial velocity of the rock?

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Final answer:

The initial velocity of the rock is a vector composed of horizontal and vertical components, calculated using trigonometry with the given speed and angle. The components are 9.0 m/s in the horizontal direction and 12.0 m/s in the vertical direction.

Step-by-step explanation:

The question is asking us to calculate the initial velocity of a rock that has been projected from ground level. The initial speed of the rock is 15.0 m/s and the launch angle is 53.0° above the horizontal. The initial velocity of a projectile thrown at an angle is a vector quantity with both horizontal and vertical components.

To find these components, we use the trigonometric functions sine and cosine along with the given speed and launch angle:

  • Horizontal component (vx0): v0 × cos(θ) = 15.0 m/s × cos(53.0°)
  • Vertical component (vy0): v0 × sin(θ) = 15.0 m/s × sin(53.0°)

To compute the accurate values:

  1. Calculate the horizontal component: vx0 ≈ 15.0 m/s × 0.6 ≈ 9.0 m/s
  2. Calculate the vertical component: vy0 ≈ 15.0 m/s × 0.8 ≈ 12.0 m/s

The initial velocity of the rock is therefore a vector composed of these two components, typically represented as v0 = (9.0 i, 12.0 j) m/s, where i and j are the unit vectors in the horizontal and vertical directions respectively.

User Mitesh Khatri
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