$\text {Horizontal} ( +ve) \hspace{1cm} \text{Vertical} ( +ve)$ $\text {Initial velocity } \hspace{1cm} V_H=12 \sin60 m/s \hspace{1cm} u_V=12 \cos60 m/s \ \text {Acceleration } \hspace{1cm} a_H= 0 \hspace{1cm} a_V=g=9.81m/s^2 \ \text {Displacement } \hspace{1cm} x=11.3m \hspace{1cm} y=H \ \text {Final velocity } \hspace{1cm} V_H=12 \sin60 m/s \hspace{1cm} v_V=? \ \text {Time} \hspace{1cm} t=? \hspace{1cm} t=?$ Horizontal motion:- Using $s=ut+\dfrac12 at^2,$ we get, $$ x=V_H t \hspace{1cm} [\because a_H=0]$$

$\therefore t = \dfrac x{V_H} =\dfrac {11.3}{12\sin60} \\ \therefore t=1.087s$

Vertical motion:-

Using $s=ut+\dfrac12 at^2,$ we get,

$$ H=u_V t + \dfrac12 gt^2=(12\cos60)(1.087)+\dfrac 12 (9.81) (1.087)^2$$

$\therefore H=12.323m \Rightarrow $ Ans