Assuming the accleration applied was constant, we have
[tex]v=v_0+at\implies v_0+20.0\,\dfrac{\mathrm m}{\mathrm s}=v_0+a(0.10\,\mathrm s)[/tex]
[tex]\implies20.0\,\dfrac{\mathrm m}{\mathrm s}=a(0.10\,\mathrm s)[/tex]
[tex]\implies a=200\,\dfrac{\mathrm m}{\mathrm s^2}[/tex]
Then the force applied to the ball is given by
[tex]F=ma=(0.45\,\mathrm{kg})\left(200\,\dfrac{\mathrm m}{\mathrm s^2}\right)[/tex]
[tex]\implies F=90\,\dfrac{\mathrm{kg}\,\mathrm m}{\mathrm s^2}=90.\,\mathrm N[/tex]
