Solution
The equations in the x and y component
[tex]\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=\frac{7}{20\cos(40^0)} \\ \\ \text{ for y component,} \\ y(t)=v_0t\sin\theta-\frac{1}{2}gt^2 \\ \\ \Rightarrow y(\frac{7}{20\cos(40^{0})})=20\times\frac{7}{20\cos(40^{0})}\times\sin(40^0)-\frac{1}{2}\times32.1741\times(\frac{7}{20\cos(40^{0})})^2 \\ \\ \Rightarrow y(\frac{7}{20\cos(40^0)})=2.5\text{ ft} \\ \end{gathered}[/tex]
Therefore, the height of the ball when it's is 7 feet away horizontally is: 7 + 2.5 = 9.5 feet
