Let [tex]\mathbf r_A[/tex] denote the position vector of the ball hit by player A. Then this vector has components
[tex]\begin{cases}r_{Ax}=\left(2.4\,\frac{\mathrm m}{\mathrm s}\right)t\\r_{Ay}=1.2\,\mathrm m-\frac12gt^2\end{cases}[/tex]
where [tex]g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}[/tex] is the magnitude of the acceleration due to gravity. Use the vertical component [tex]r_{Ay}[/tex] to find the time at which ball A reaches the ground:
[tex]1.2\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.49\,\mathrm s[/tex]
The horizontal position of the ball after 0.49 seconds is
[tex]\left(2.4\,\dfrac{\mathrm m}{\mathrm s}\right)(0.49\,\mathrm s)=12\,\mathrm m[/tex]
So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector [tex]\mathbf r_B[/tex] of the ball hit by player B has
[tex]\begin{cases}r_{Bx}=v_0t\\r_{By}=1.6\,\mathrm m-\frac12gt^2\end{cases}[/tex]
Again, we solve for the time it takes the ball to reach the ground:
[tex]1.6\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.57\,\mathrm s[/tex]
After this time, we expect a horizontal displacement of 12 meters, so that [tex]v_0[/tex] satisfies
[tex]v_0(0.57\,\mathrm s)=12\,\mathrm m[/tex]
[tex]\implies v_0=21\,\dfrac{\mathrm m}{\mathrm s}[/tex]
