Question

dana shoots a basketball at an angle of 45° from the horizontal. it leaves her hands 6 feet from the ground with a velocity of 27 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: Parametric equation for horizontal position

The general formula for the horizontal position \(x(t)\) in projectile motion is \(x(t)=(v\cos\theta)t\), where \(v\) is the initial velocity and \(\theta\) is the angle of projection.
Given \(v = 27\) ft/s and \(\theta=45^{\circ}\), and \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\approx0.7\).
So, \(x(t)=27\times\cos45^{\circ}\times t\approx 19.1t\)

Step2: Parametric equation for vertical position

The general formula for the vertical position \(y(t)\) in projectile motion is \(y(t)=- 16t^{2}+(v\sin\theta)t + h\), where \(v\) is the initial velocity, \(\theta\) is the angle of projection, \(h\) is the initial height.
Given \(v = 27\) ft/s, \(\theta = 45^{\circ}\), \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\approx0.7\), and \(h = 6\) feet.
So, \(y(t)=-16t^{2}+27\times\sin45^{\circ}\times t + 6\approx-16t^{2}+19.1t + 6\)

Answer:

The parametric equations are \(x(t)\approx19.1t\) and \(y(t)\approx - 16t^{2}+19.1t + 6\)