A ball is dropped from the top of a tall building of height = 50 m. About how long does it take for the ball to hit the ground?(Neglect air...

Let the acceleration due to gravity at place be g = 9.8
m/s^2.

The motion under gravitation of the ball dropped
from the top of the guilding is given by:

s(t) = (1/2)gt^2,
where s is the vertically downward distance traveeled by the  ball from the top of the
building in time t seconds.

S the time to reach the ground
t is given by the equation when s(t) = 50m.

So, 50 =
(1/2)gt^2. g= 9.8m/s assumed.

 t^2 =
2*50/g

t = sqrt(100/9.8) = 3.1944 seconds
approximately.

So the ball takes 3.1944
seconds.

To find the momentum of the ball when it hits the
ground.

Momentum = mv , where m is the mass of the ball
 * v velocity of the  ball when it hits the
ground.......(1)

So the velocity of the ball when it
strikes the ground = u +gt , where u = 0 the initial velocity when the ball was
dropped.

 v = gt = 9.8* sqrt(100/9.8) =
31.304951695

When the ball recoils it has only half the
momentum . So the ball has an initial velocity of  31.3045/2 = 15.65247584
m/s.

So the again the ball follows the law of
motion:

 Smax = (v^2-u^2)/2g, where u = initial velocity =
15.6524...m/s as found above.  v = final velocity when the ball has 0 velocity and
starts returning to ground reaching the maximum height
Smax.

So  Smax = (0-15.65247584^2)/(2*-9.8) = 12.5
meter.

Therefore , the ball with half the momentum reach 
12.5 meter of maximum height (that is quarter of the height of the
building.)