Assume the ball was thrown at ground level, so its initial position vector is [tex]\mathbf r(0)=\mathbf 0[/tex]. Let [tex]v_0[/tex] denote the ball's initial speed and [tex]\theta[/tex] the angle at which the ball is thrown.
Over time, its position changes according to [tex]\mathbf r(t)=x\,\mathbf i+y\,\mathbf j[/tex], where the components [tex]x,y[/tex] are determined by
[tex]x=v_0\cos\theta\,t[/tex]
[tex]y=v_0\sin\theta\,t-\dfrac g2t^2[/tex]
where [tex]g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}[/tex] is the acceleration due to gravity. We know the ball is in the air for 4.5 seconds and that it travels a horizontal distance of 45 meters, which means
[tex]\begin{cases}45\,\mathrm m=v_0\cos\theta(4.5\,\mathrm s)\\0=v_0\sin\theta(4.5\,\mathrm s)-\frac g2(4.5\,\mathrm s)^2\end{cases}\implies\begin{cases}v_0\cos\theta=10\,\frac{\mathrm m}{\mathrm s}\\v_0\sin\theta=22.05\,\frac{\mathrm m}{\mathrm s}\end{cases}[/tex]
Divide the second equation by the first to eliminate [tex]v_0[/tex], and we can solve for [tex]\theta[/tex], expecting some solution within [tex]0<\theta<90^\circ[/tex]:
[tex]\dfrac{v_0\sin\theta}{v_0\cos\theta}=\tan\theta=2.205\implies\theta\approx66^\circ[/tex]
Then plugging this into either equation above will tell us the ball's initial speed:
[tex]v_0\cos66^\circ=10\,\dfrac{\mathrm m}{\mathrm s}\implies v_0\approx25\,\dfrac{\mathrm m}{\mathrm s}[/tex]
