Let's make a diagram to visualize the problem.
Given that the ball is thrown upwards, there will be a point where the ball will stop and go back to reach the ground. At this maximum point, the velocity is zero.
[tex]v_b=0[/tex]We know by given that the initial velocity is 38 m/s (at point A).
[tex]v_a=38(\frac{m}{s})[/tex]To find the time taken by the ball to be found 265 meters above the ground, we have to use the following formula.
[tex]h=v_0t+\frac{1}{2}gt^2[/tex]Where the initial velocity is 38 m/s, the gravity is 9.8 m/s^2, and the height is 25m because that's the position from the top of the building.
[tex]\begin{gathered} 25=38t+\frac{1}{2}\cdot9.8t^2 \\ 0=-25+38t+4.9t^2 \end{gathered}[/tex]Let's use the quadratic formula to find the solution to the quadratic equation.
[tex]t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Where a = 4.9, b = 38, and c = -25.
[tex]\begin{gathered} t=\frac{-38\pm\sqrt[]{38^2-4\cdot4.9\cdot(-25)}}{2(4.9)}=\frac{-38\pm\sqrt[]{1444+490}_{}}{9.8} \\ t=\frac{-38\pm\sqrt[]{1934}_{}}{9.8}\to t_1\approx0.61\to t_2\approx-8.37 \end{gathered}[/tex]To find the velocity of the ball the moment it reaches the ground, we have to find, first, the maximum height reached.
[tex]v^2_f=v^2_0+2gh_{\max }[/tex]Where the final velocity is null because we are calculating the maximum height.
[tex]\begin{gathered} 0=(38)^2+2(-9.8)h_{\max } \\ 0=1444-19.6h_{\max } \\ -1444=-19.6h_{\max } \\ h_{\max }=\frac{-1444}{-19.6} \\ h_{\max }=73.67m \end{gathered}[/tex]Then, we sum the maximum height with the height of the building to find the total distance from point B to point C.
[tex]h_{BC}=73.67m+240m=313.67m[/tex]Now we are able to find the final velocity of the ball when it reaches the ground.
[tex]\begin{gathered} v^2_f=v^2_0+2gh_{BC} \\ v^2_f=0+2(9.8)(313.67)_{} \\ v_f=\sqrt[]{6147.93} \\ v_f\approx78.41(\frac{m}{s}) \end{gathered}[/tex]
