Taking vertical downward motion of ball from a height 90 m, wehave
u=0,a=10 m∕ s2;
S=90 m,t=?;v=?
t=√

2S

a

=√

2×90

10

=3√2 s=4.24 s
v=√2aS
=√2×10×90
=30√2 m∕ s
Rebound velocity of ball,
u,=

9

10

v
=

9

10

×30√2
=27√2 m∕ sTime to reach the highest point is,
t,=

u,

a

=

27√2

10

=2.7√2=3.81 s
Total time=t+t′=4.24+3.81=8.05sThe ball will take further 3.81 s to fall back to floor, where its velocity before striking the floor =27√2 m∕s.
Velocity of ball after striking the floor
=

9

10

×27√2
=24.3√2 m∕ s
Total time elapsed before upwardmotion of ball =8.05+3.81=11.86s
Thus the speed-time graph of thismotion will be as shown in figure.