#### 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 $v_0$ be the speed at the point of release.
Let $y$ be the maximum height.
$y = \frac{v_0^2}{2g}$
Let $v$ be the speed 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 $t_1$ be the time for the object to rise to its maximum height $y$.
$t_1 = \frac{v_0}{g}$
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.