Given,
Distance travelled, d=44 m.
The time of flight t=2.9 s
The angle is
[tex]\theta=45^o[/tex]
Let the initial velocity be v
The motion is projectile.
Thus the horizontal distance travelled is
[tex]\begin{gathered} d=(v\cos \theta)t \\ \Rightarrow44=v\times\cos 45^o\times2.9 \\ \Rightarrow v=21.5\text{ m/s} \end{gathered}[/tex]
Thus the answer is 21.5 m/s
