dana shoots a basketball at an angle of 45° from the horizontal. it leaves her hands 6 feet from the ground…
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.
Answer
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)