Question

A computer rando...Student Dashboard...002 > Modules > Hawkes > Hawkes Single Sign OnMCSave & Exit Certify Lesson: 9.4 Parametric Equations

Answer 1

The equations in the x and y component

`\begin{gathered} x(t)=v_0t\cos\theta \\ \Rightarrow x(t)=20t\cos(40^0) \\ \text{ when }x(t)=7 \\ \Rightarrow7=20t\cos(40^0) \\ \Rightarrow t=(7)/(20\cos(40^0)) \\ \\ \text{ for y component,} \\ y(t)=v_0t\sin\theta-(1)/(2)gt^2 \\ \\ \Rightarrow y((7)/(20\cos(40^(0))))=20*(7)/(20\cos(40^(0)))*\sin(40^0)-(1)/(2)*32.1741*((7)/(20\cos(40^(0))))^2 \\ \\ \Rightarrow y((7)/(20\cos(40^0)))=2.5\text{ ft} \\ \end{gathered}`

Therefore, the height of the ball when it's is 7 feet away horizontally is: 7 + 2.5 = 9.5 feet