We will have the following:
a)
[tex]\begin{gathered} x_0=0m \\ \\ y_0=49m \end{gathered}[/tex]
b)
[tex]\begin{gathered} v_{0x}=19m/s \\ \\ v_{0y}=0m/s \end{gathered}[/tex]
c)
[tex]\begin{gathered} v_x=v_{ix}+a_xt\Rightarrow v_x=19m/s+(0m/s^2)t\Rightarrow v_x=19m/s \\ \\ v_y=v_{iy}+a_yt\Rightarrow v_y=0m/s+(9.8m/s^2)t\Rightarrow v_y=(9.8m/s^2)t \end{gathered}[/tex]
d)
[tex]\begin{gathered} x=v_{ix}t+\frac{1}{2}a_xt^2\Rightarrow x=19m/s+\frac{1}{2}(0m/s^2)t^2\Rightarrow x=(19m/s)t \\ \\ y=v_{iy}t+\frac{1}{2}a_yt^2\Rightarrow y=(0m/s)t+\frac{1}{2}(9.8m/s^2)t^2\Rightarrow y=\frac{1}{2}(9.8m/s^2)t^2 \end{gathered}[/tex]
e)
[tex]\begin{gathered} 49m=\frac{1}{2}(9.8m/s^2)t^2\Rightarrow t=\sqrt{\frac{49m}{4.9m/s^2}} \\ \\ \Rightarrow t=\sqrt{10}s\Rightarrow t\approx3.2s \end{gathered}[/tex]
So, the time it takes is sqrt(10) s, that is approximately 3.2s.
f)
[tex]\begin{gathered} v_f=\sqrt{(19m/s)^2+((9.8m/s^2)(\sqrt{10}s))^2}\Rightarrow v_f=36.35106601...m/s \\ \\ \Rightarrow v_f\approx36.4m/s \end{gathered}[/tex]
So, the final velocity is approximately 36.4 m/s.
[tex]\begin{gathered} 19m/s=(36.4m/s)cos(\theta)\Rightarrow cos(\theta)=\frac{19m/s}{36.4m/s} \\ \\ \Rightarrow\theta=cos^{-1}(\frac{19}{36.4})\Rightarrow\theta=58.53497293... \\ \\ \Rightarrow\theta\approx58.5 \end{gathered}[/tex]
So, the angle would be approximately 58.5°.
