Answer:
Distance of 1.956 m
Explanation:
Consider the vertical plane
[tex]S=ut+\frac{1}{2} at^2[/tex]
where u=o, S=4.5-15=3 m
[tex]3=0\times t+\frac{1}{2} 9.8 t^2\\t=\sqrt{0.612} \\t=0.782 s[/tex]
now consider horizontal plane
u=2.5 m/s, a=0
[tex]S=2.5\times 0.782+0\\S=1.956 m[/tex]