Answer:
The answer is "[tex]16.79\ \frac{m}{s}[/tex]"
Explanation:
In this question, the halfway indicates the height that is [tex]\frac{28.8}{2}=14.4 \ m[/tex]
Using formula:
[tex]v^2=u^2+2as\\\\v^2=24^2+2(-10)(14.4)\\\\[/tex]
[tex]v^2=576-288\\\\v^2=288\\\\v=\sqrt{288}\\\\v=16.97 \ \frac{m}{s}[/tex]
