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A ball is thrown at an angle of 30° off the horizontal, with an initialvelocity of 28 m/s. What is the maximum height the ball will reach?

Answer :

[tex]sen(30)\ast28=\frac{14m}{s}[/tex][tex]\begin{gathered} v^2=(14m/s)^2-2ad \\ \frac{14^2}{2\ast9.81}=d=9.99m \end{gathered}[/tex]

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