Answer :
Given:
The initial speed of the object is,
[tex]u=20\text{ m/s}[/tex]The maximum height is,
[tex]H=12\text{ m}[/tex]To find:
The angle of launch above the X-axis
Explanation:
The maximum height above the X-axis is,
[tex]H=\frac{u^2sin^2\theta}{2g}[/tex]Here, the angle above the X-axis is,
[tex]\theta[/tex]Substituting the values we get,
[tex]\begin{gathered} 12=\frac{(20)^2sin^2\theta}{2\times9.8} \\ sin^2\theta=\frac{12\times2\times9.8}{400} \\ sin\theta=0.7668\text{ \lparen taking positive value only\rparen} \\ \theta=50.1\degree \end{gathered}[/tex]Hence, the angle of launch is,
[tex]50.1\degree[/tex]