horizontal distance = 5 meteres
height = 1.5 meters
Therefore,
y component
[tex]\begin{gathered} 0=1.5+\frac{1}{2}(-9.8)t^2 \\ 1.5=4.9t^2 \\ t^2=\frac{1.5}{4.9} \\ t=\sqrt[]{0.30612244898} \\ t=0.55328333517 \\ t=0.55\text{ s} \end{gathered}[/tex]The maximum speed(horizontal) can be computed as follows
[tex]\begin{gathered} S_x=ut \\ 5=0.55u \\ u=\frac{5}{0.55} \\ u=9.09090909091 \\ u=9.09\text{ m/s} \end{gathered}[/tex]The maximum speed is the initial horizontal velocity of the x-axis.
