Given
Let the initial speed is V
Angle at which rubber band is launched = 32.5
Horizontal component of initial velocity
[tex]\begin{gathered} v_x=v\cos \theta \\ \end{gathered}[/tex]Time is given as t = 1.60 sec
Distance in horizontal direction = 1.30 m
Procedure
[tex]\begin{gathered} t=\frac{x}{v_x} \\ t=\frac{x}{v\cos \theta} \\ v=\frac{x}{t\cos \theta} \\ v=\frac{1.30m}{1.60s\cdot\cos 32.5} \\ v=0.9633\text{ m/s} \end{gathered}[/tex]The answer would be v = 0.9633 m/s
