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Rick shoots a basketball at an angle of 60' from the horizontal. It leaves his hands 6 feet from the ground with a velocity of 25 ft/s.Step 1 of 2: Construct a set of parametric equations describing the shot. Answer

Rick shoots a basketball at an angle of 60' from the horizontal. It leaves his hands-example-1

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Solution:

Given:


\begin{gathered} Initial\text{ velocity,}u=25ft\text{ /s} \\ \theta=60^0 \end{gathered}

The parametric equations are gotten by first resolving the velocity into horizontal and vertical components.

Recall;


\begin{gathered} speed=(distance)/(time) \\ distance=speed* time \end{gathered}

Hence, the parametric equations are:


\begin{gathered} x=(25cos60)t \\ y=(25sin60)t+6 \end{gathered}

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