Answer
(a) $0.49~s$
(b) $4.8~m/s$
Work Step by Step
Let the positive direction be downward and the origin be the starting position of the book.
(a) $x_{0book}=0$, $x_{0floor}=1.2~m$, $v_{0~book}=v_{0~floor}=3.0~m/s$, $a_{book}=g=9.81~m/s^2$, $a_{floor}=0$
$x=x_0+v_0t+\frac{1}{2}at^2$
$x_{book}=x_{0~book}+v_{0~book}t+\frac{1}{2}a_{book}t^2$
$x_{book}=0+(3.0~m/s)t+\frac{1}{2}(9.81~m/s^2)t^2$
$x_{book}=(3.0~m/s)t+(4.905~m/s^2)t^2$
$x_{floor}=x_{0floor}+v_{0~floor}t+\frac{1}{2}a_{floor}t^2$
$x_{floor}=1.2~m+(3.0~m/s)t+\frac{1}{2}0t^2$
$x_{floor}=1.2~m+(3.0~m/s)t$
When the book reaches the floor $x_{book}=x_{floor}$:
$(3.0~m/s)t+(4.905~m/s^2)t^2=1.2~m+(3.0~m/s)t$
$(4.905~m/s^2)t^2=1.2~m$
$t=\sqrt {\frac{1.2~m}{4.905~m/s^2}}=0.49~s$
(b) $v=v_{0}+at$
$v_{book}=v_{0~book}+at$
$v_{book}=3.0~m/s+(9.81~m/s^2)(0.49~s)=7.8~m/s$
My speed is constant and equal to the speed of the elevator ($3.0~m/s$).
Let $v_{r}$ be the book's speed relative to me.
$v_{r}=7.8~m/s-3.0~m/s=4.8~m/s$
