Question

rick shoots a basketball at an angle of 35° from the horizontal. it leaves his hands 7 feet from the ground with a velocity of 20 ft/s. step 1 of 2: construct a set of parametric equations describing the shot. round all final values to the nearest tenth.

Explanation:

Step1: Find the horizontal component of velocity

The horizontal component of velocity \(v_x\) is given by \(v\cos\theta\), where \(v = 20\) ft/s and \(\theta=35^{\circ}\).
\(v_x=20\cos(35^{\circ})\approx20\times0.819\approx16.4\) ft/s.
The parametric equation for the horizontal position \(x(t)\) is \(x(t)=v_x t\), so \(x(t)=16.4t\).

Step2: Find the vertical component of velocity and initial height

The vertical component of velocity \(v_y\) is given by \(v\sin\theta\), where \(v = 20\) ft/s and \(\theta = 35^{\circ}\).
\(v_y=20\sin(35^{\circ})\approx20\times0.574\approx11.5\) ft/s.
The initial height \(y_0 = 7\) feet.
The parametric equation for the vertical position \(y(t)\) is \(y(t)=- 16t^{2}+v_y t + y_0\).
Substituting the values, we get \(y(t)=-16t^{2}+11.5t + 7\).

Answer:

The parametric equations are \(x(t)=16.4t\) and \(y(t)=-16t^{2}+11.5t + 7\)