Answer
(a) The object will have the same speed when it returns to the point of release.
(b) The time of flight will be twice the time to get to the highest point.
Work Step by Step
(a) Let $v_i=v_0$ be the speed at the point of release and $v_f=0$ be the speed at final point at maximum height.
By using equation of motion
$2ha={v_f^2}-{v_i^2}$ (1)
Let $h=y$ be the maximum height and $a=-g$.
By using values in (1) we can have
$y = \frac{v_0^2}{2g}$
Let $v_i=0$ be the speed at starting point when starts moving dwon and $v_f=v$ be the speed at the point when the object returns to the point of release.
$v^2 = 0 + 2gy = 2g(\frac{v_0^2}{2g}) = v_0^2$
$v = v_0$
The object will have the same speed when it returns to the point of release.
(b) Let $t=t_1$ be the time for the object to rise from initial point where $v_i=0$ to its maximum height $y$ where $v_f=0$.
Now by using equation
$v_f=v_i+at$(2)
As $a=-g$
By using value in (2) we have
$t_1 = \frac{v_0}{g}$
Let $t_2$ be the time for the object to fall back down to the point of release.
$t_2 = \frac{v}{g}$
Let $t$ be the total time of flight.
$t = t_1+t_2 = \frac{v_0}{g} + \frac{v}{g}$
$t = \frac{v_0}{g}+\frac{v_0}{g} = 2\times \frac{v_0}{g}$
$t = 2\times t_1$
The time of flight will be twice the time to get to the highest point.
