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You are fixing the roof of your house when a hammer breaks loose and slides down. The roof makes an angle of 65o∘ with the horizontal, and the hammer is moving at 9.5 m/s when it reaches the edge. Assume that the hammer is moving from the top of the roof to its right edge.

What is the horizontal component of the hammer's velocity just as it leaves the roof?
Express your answer with the appropriate units. Enter positive value if the x-component of the velocity is to the right and negative value if the x-component of the velocity is to the left.
What is the vertical component of the hammer's velocity just as it leaves the roof?
Express your answer with the appropriate units. Enter positive value if the direction of the y-component of the velocity is upward and negative value if the y-component of the velocity is downward.

Respuesta :

Answer:

[tex]v_x\approx4.0149\ m.s^{-1}[/tex]

[tex]v_y\approx-8.6099\ m.s^{-1}[/tex]

Explanation:

Given:

initial speed of the hammer when leaving the edge of the roof along the inclination of the roof, [tex]v=9.5\ m.s^{-1}[/tex]

inclination of the roof form horizontal, [tex]\theta=65^{\circ}[/tex]

  • Since the hammer is moving from the top of the roof to the right edge, its horizontal component will be towards right and vertical component will be towards downward direction.

Now the horizontal velocity:

[tex]v_x=v.\cos\theta[/tex]

[tex]v_x=9.5\times \cos65^{\circ}[/tex]

[tex]v_x\approx4.0149\ m.s^{-1}[/tex]

The vertical velocity:

[tex]v_y=-v.\sin\theta[/tex]

[tex]v_y=-9.5\times \sin65^{\circ}[/tex]

[tex]v_y\approx-8.6099\ m.s^{-1}[/tex]

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