Answer :
Given,
The range of the projectile, R=43.0 m
The angle at which the bucket is projected, θ=72.0°
The range of a projectile is the maximum displacement of the projectile in the x-direction.
And the range is given by,
[tex]R=\frac{u^2\sin (2\theta)}{g}[/tex]Where u is the initial velocity and g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} 43.0=\frac{u^2\times\sin (2\times72.0^{\circ})}{9.8} \\ u=\sqrt{\frac{43\times9.8}{\sin(2\times72.0^{\circ})}} \\ =26.77\text{ m/s} \end{gathered}[/tex]Thus the initial velocity must be 26.77 m/s