Answer:2.45 m/s
Explanation:
Given
Launch velocity[tex](u)=3 m/s[/tex]
launch angle[tex]=35^{\circ}[/tex]
as the vertical velocity first decreasing to zero and then increases to original value so its avg is zero .
[tex]v_{avg}=\frac{displacement}{time}[/tex]
[tex]v_{avg}=\frac{Range}{time}[/tex]
[tex]v_{avg}=\frac{u\cos \theta \times t}{t}[/tex]
thus [tex]v_{avg}=\frac{3\cos 35\times t}{t}[/tex]
[tex]v_{avg}=3\cos 35=2.45 m/s[/tex]
