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A projectile is fired from the top of a 300 m tall building with an initial velocity of Vi= 85.0 m/s and an initial angle theta i=65.0° above the horizontal. Answer the following questions. [Use g = 9.80 m/s]Question 9What is the x-component of the initial velocity?Round your answer to 3 significant figures.Question 10What is the y-component of the initial velocity?Round your answer to 3 significant figures.Add your answer

User Ludwiguer
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1 Answer

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In order to determine the x and y components of the initial velocity, use the following formulas:


\begin{gathered} v_(ox)=v_o\cos \theta \\ v_(oy)=v_o\sin \theta \end{gathered}

where,

vo: magnitude of the initial velocity = 85.0 m/s

vox: x component of vo = ?

voy: y component of vo = ?

θ: angle above the horizontal = 65.0°

Replace the previous values into the expressions for vox and voy:


\begin{gathered} v_(ox)=(85.0(m)/(s))\cos (65.0)=35.9(m)/(s) \\ v_(oy)=(85.0(m)/(s))\sin (65.0)=77.03(m)/(s) \end{gathered}

Hence, you obtain:

vox = 35.9 m/s

voy = 77.0 m/s

User Crystark
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