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
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Solution:
Given:
[tex]\begin{gathered} Initial\text{ velocity,}u=25ft\text{ /s} \\ \theta=60^0 \end{gathered}[/tex]
The parametric equations are gotten by first resolving the velocity into horizontal and vertical components.
Recall;
[tex]\begin{gathered} speed=\frac{distance}{time} \\ distance=speed\times time \end{gathered}[/tex]Hence, the parametric equations are:
[tex]\begin{gathered} x=(25cos60)t \\ y=(25sin60)t+6 \end{gathered}[/tex]