Answer:
[tex]V=2.8534m/s \approx 3m/s[/tex]
Explanation:
From the question we are told that
Acceleration [tex]a=0.460m/s[/tex]
Distance traveled [tex]S_1=8.5m[/tex]
Distance traveled [tex]S_2=15.6m[/tex]
Generally the velocity of the ball at the bottom of the first plane V is mathematically given by
[tex]v^2=2as[/tex]
[tex]v=\sqrt{2as}[/tex]
[tex]V=\sqrt{2*0.460*8.85}[/tex]
[tex]V=2.8534m/s \approx 3m/s[/tex]