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Answer :
Let's make a diagram to visualize the problem.
It's important to know that this motion is not like a parabola because the ball is thrown upwards. First, we find the initial velocity.
[tex]h=v_0t+\frac{1}{2}gt^2[/tex]Using the given magnitudes, we have the following
[tex]\begin{gathered} 53.9=v_0\cdot4.15+\frac{1}{2}\cdot(-9.8)\cdot(4.15)^2 \\ 53.9=4.15v_0-84.40 \\ 53.9+84.40=4.15v_0 \\ v_0=\frac{138.3}{4.15}(\frac{m}{s}) \\ v_0\approx33.33(\frac{m}{s}) \end{gathered}[/tex]The initial velocity is 33.33 m/s.
Now we are able to find the final velocity of the ball.
[tex]\begin{gathered} v_f=v_0+gt \\ v_f=33.33-9.8\cdot4.15 \\ v_f=33.33-40.67 \\ v_f=-7.34(\frac{m}{s}) \end{gathered}[/tex]Therefore, the speed of the ball when it lands on the roof of the adjacent building is 7.43 m/s, going downwards.
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Rewritten by : Barada
