Let [tex]v_0[/tex] be the initial velocity of the ball. Then it has components
[tex]\begin{cases}v_{0x}=|v_0|\cos35^\circ\\v_{0y}=|v_0|\sin35^\circ\end{cases}[/tex]
The position vector [tex](x,y)[/tex] of the ball at time [tex]t[/tex] has components
[tex]\begin{cases}x=v_{0x}t\\y=1.0\,\mathrm m+v_{0y}t-\frac g2t^2\end{cases}[/tex]
where [tex]g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}[/tex] is the acceleration due to gravity. When the ball is 90.0 m away from home plate, we're told that it clears (presumably) a barrier of some kind that is 7.50 m tall, so that for some time [tex]t[/tex] we get
[tex]\begin{cases}90.0\,\mathrm m=v_{0x}t\\7.50\,\mathrm m=1.0\,\mathrm m+v_{0y}t-\frac g2t^2\end{cases}[/tex]
Solving for this [tex]t[/tex] gives
[tex]t=\dfrac{90.0\,\mathrm m}{v_{0x}}[/tex]
Substituting this into the second equation gives
[tex]6.50\,\mathrm m=v_{0y}\left(\dfrac{90.0\,\mathrm m}{v_{0x}}\right)-\dfrac g2\left(\dfrac{90.0\,\mathrm m}{v_{0x}}\right)^2[/tex]
[tex]6.50\,\mathrm m=90\tan35^\circ-\dfrac{(90.0\,\mathrm m)^2g}{2|v_0|^2\cos^235^\circ}[/tex]
[tex]\implies|v_0|^2=1046.55\,\dfrac{\mathrm m^2}{\mathrm s^2}[/tex]
[tex]\implies|v_0|=32.4\,\dfrac{\mathrm m}{\mathrm s}[/tex]
