An object is projected vertically upward from the ground. Show that it takes the object the same amount of time to reach its maximum height as...

Let u be the initial velocity of the object  thrown up
verically.

So the ovject goes on loosing its velocity till the final
velocity of the object becomes zero.

Therefore  v = u-gt = 0, where
v is the final velocity and g is the acceleration due to the gravity. So u = gt or time when the
velocity becomes zero is u/g.

Therefore the height reached by the
object h = ut -(1/2)gt^2. Substituting t = u/g in this equation h = u(u/g)-(1/2)g(u/g)^2 =
u^2/g-u^2/2g = u^2/2g. Therefore the maximum height reached by the object is u^2/2g in time t =
u/g.

Now we calculate the time required for the object to fall from
the height  u^2/2g to the ground . Let t be the time to reach the
ground.

Then the equation of motion is h = u^2/g = u't+(1/2)gt^2, 
where u' = is the initial velocity = 0 when the object has reached the highest point and lost all
its velocity and begins fall.

u^2/2g = 0*t +(1/2)gt^2 . We solve for
t. We multiply both sides by g/2:

(2/g) *u^2/2g =
t^2.

u^2/g^2 = t^2.

Therefore u/g = t.
Or the time for the object to reach  the ground from the highest point is u/g which is the same
as the time to reach the highest point from the ground.