We know that
• The mass is 11.54 kg.
,• The time taken is 0.651 seconds.
,• The distance covered is 22.87 meters.
,• The resutant force is 236.55 N.
,• The angle of launch is 57.66.
Let's use Newton's Second Law.
[tex]\Sigma F_y=ma\to F-W=ma[/tex]Where F = 236.55N, W = mg, m = 11.54kg, and g = 9.81 m/s^2.
[tex]\begin{gathered} 236.55N-11.54\operatorname{kg}\cdot9.81\cdot\frac{m}{s^2}=11.54\operatorname{kg}\cdot a \\ 236.55N-113.21N=11.54\operatorname{kg}\cdot a \\ a=\frac{123.34N}{11.54\operatorname{kg}} \\ a\approx10.7\cdot\frac{m}{s^2} \end{gathered}[/tex]The acceleration of the object is 10.7 m/s^2.
Now, let's find the initial vertical velocity.
[tex]\begin{gathered} v_y=v_{oy}+gt \\ 0=v_{oy}-9.8\cdot0.651\sec \\ v_{0y}=6.4\cdot\frac{m}{s} \end{gathered}[/tex]