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Question

A ball is shot from the top of a building with an initial velocity of 24 m/s24 \mathrm{~m} / \mathrm{s} at an angle θ=42\theta=42^{\circ} above the horizontal. What are the horizontal and vertical components of the initial velocity?

Solution

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Answered 1 year ago
Answered 1 year ago
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Given:

A ball being shot with the following specifications:

  • It is shot from the top of a certain building.
  • It is shot with an initial velocity of 24 m/s24\text{ m}/\text{s}.
  • The launching angle is θ=42°\theta=42\degree.

Required:

  • To determine the components of the initial velocity in the horizontal direction and the vertical direction.

Context:

The components of the initial velocity are given by:

vx0=v0cos(θ)(1)v_{x0}=v_0\cdot \cos(\theta)\tag 1

vy0=v0sin(θ)(2)v_{y0}=v_0\cdot \sin(\theta)\tag 2

Where:

  • vx0v_{x0} is the horizontal component of the initial velocity vector.
  • vy0v_{y0} is the vertical component of the initial velocity vector.
  • v0=24 m/sv_0=24\text{ m}/\text{s} is the initial speed of the ball.

We only need to use Eqs. (1) and (2) to determine what's being requested.

Note: Since all the information in this problem is provided using 2 significant figures, the final results will be presented using the said amount of significant figures; however, for intermediate calculations, we may use up to 3 decimal places depending on the case, even if it means using more than 2 significant figures for those intermediate instances.

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