Answer:
(A): Using the equations of motion, we can determine the answer as follows:
[tex]\begin{gathered} x(t)=x_{\circ}+v_{\circ}cos(\theta)t\rightarrow(1) \\ \\ y(t)=y_{\circ}+v_{\circ}sin(\theta)-\frac{1}{2}gt^2\rightarrow(2) \\ \\ y(x)=xtan(\theta)-\frac{g}{2(v_{\circ})^2cos^2(\theta)}x^2\rightarrow(3) \end{gathered}[/tex]
formula (3) is obtained from (1) and (2), using equation (3) the answer is determined as below:
[tex]\begin{gathered} y(x)=xtan(\theta)-\frac{g}{2(v_{\circ})^2cos^2(\theta)}x^2 \\ \\ v_{\circ}=22\text{ f/s} \\ \\ \theta=45 \\ \\ g=32.1522\text{ f/s} \\ \\ y(x)=xtan(45)-\frac{32.1522}{2\times22^2cos^2(45)}x^2 \\ \\ y(x)=x-\frac{32.1522}{2\times22^2cos^2(45)}x^2 \\ \\ y(x)=x-\frac{32.152,2}{484}x^2 \\ \\ y(x)=x-0.06643x^2 \\ \\ (x,y)\rightarrow\text{ Adjusting the position for the shift gives:} \\ \\ y(x)=[(x+10)-0.06643(x+10)^2]+6\rightarrow(4) \end{gathered}[/tex]
The plot of the (4) reveals the following:
Therefore the answer is no.
(D) Trying a new angle theta = 60 degrees gives the following new answer:
Therefore the answer is:
[tex]\theta=60^{\circ}[/tex]
