a) 13.25 m
b)22.3 degrees
Explanation
Step 1
[tex]\text{distance = velocity }\cdot time[/tex]Let
[tex]\begin{gathered} \text{distance = 38} \\ \text{time}=\text{ t} \\ \text{speed}=\text{ }23.1\frac{m}{s} \end{gathered}[/tex]so
[tex]\begin{gathered} \text{distance = velocity }\cdot time \\ 38\text{ m=23.1}\cdot t \\ t=\frac{38\text{ m}}{23.1}=1.645\text{ s} \\ \text{time}=\text{ 1.645} \end{gathered}[/tex]now , to find the disntace to the targe let's use the formula
[tex]\begin{gathered} y=V_0\sin _{}\text{ (}\propto\text{)}+\frac{1}{2}gt^2 \\ as\text{ the angle is 0 } \\ y=V_0\sin _{}\text{ (0)}+\frac{1}{2}gt^2 \\ y=\frac{1}{2}gt^2 \end{gathered}[/tex]hence
[tex]\begin{gathered} y_f=-\frac{1}{2}g\cdot t^2 \\ \text{replace} \\ y_f=-\frac{1}{2}(9.8)\cdot(1.645^2) \\ y_f=-\frac{1}{2}(9.8)\cdot(2.7) \\ y_f=-13.25 \\ \end{gathered}[/tex]so,target missed by 13.25 meters.
Step 2
b)At what angle should the bow be aimed in order to hit the target?
to solve this we can use the expression
[tex]\begin{gathered} t=\frac{dis\tan ce(x)}{velocity\text{ (x)}} \\ so \\ \text{distance}=\text{ 38 m} \\ V_x=V\text{ cos (}\propto) \\ V_x=23.1\text{ cos (}\propto) \\ \end{gathered}[/tex]now, replace and solve for the angle
[tex]\begin{gathered} t=\frac{dis\tan ce(x)}{velocity\text{ (x)}} \\ =\frac{38}{23.1\cos \propto} \\ \\ \frac{38}{23.1\cos\propto}=\frac{23.1\sin \propto}{4.9} \\ 38\cdot4.9=23.1\sin \propto\cdot23.1\cos \propto \\ \frac{38\cdot4.9}{23.1^2}=\sin c\propto\cos \propto \\ so \\ \propto=22.3 \end{gathered}[/tex]so, the angle should be
22.3 degrees
