Taking vertical downward motion of ball from a height 90 m, wehave $u=0, a=10\,\mathrm{m}/\mathrm{s}^{2};$ $S=90\,\mathrm{m}, t=? ; v=?$ $t=\sqrt{\frac{2S}{a}}$ $=\sqrt{\frac{2 \times 90}{10}}$ $=3\sqrt{2}\,\mathrm{s}=4.24\,\mathrm{s}$ $v=\sqrt{2aS}$ $=\sqrt{2 \times 10 \times 90}$ $=30\sqrt{2}\,\mathrm{m}/\mathrm{s}$ Rebound velocity of ball, $u'=\frac{9}{10}v$ $=\frac{9}{10} \times 30\sqrt{2}$ $=27\sqrt{2}\,\mathrm{m}/\mathrm{s}$Time to reach the highest point is, $t'=\frac{u'}{a}$ $=\frac{27\sqrt{2}}{10}$ $=2.7\sqrt{2}=3.81\,\mathrm{s}$ Total time$= t + t' = 4.24 + 3.81 = 8.05\,\mathrm{s}$The ball will take further 3.81 s to fall back to floor, where its velocity before striking the floor $=27\sqrt{2}\,\mathrm{m}/\mathrm{s}$.Velocity of ball after striking the floor$=\frac{9}{10} \times 27\sqrt{2}$$=24.3\sqrt{2}\,\mathrm{m}/\mathrm{s}$Total time elapsed before upwardmotion of ball $=8.05 + 3.81 = 11.86\,\mathrm{s}$Thus the speed-time graph of thismotion will be as shown in figure.