From the edge of a building, a ball is thrown upward with the
vertical initial velocity of 19.6 m/s; when the ball returns to the
lunching level (the roof of the building) it
just miss the edge of the building and it continues falling down to the
street level. Calculate the displacement and velocity of the ball 0s,
1s, 2s, 3s, 4s, and 5s after the ball is originally thrown.

The table above can be filled in with the results of
your own calculation. However, if a wrong input is entered in a cell,
the table will indicate so by displaying a warning in the cell. The word
"Wrong" will be appended to your number. Complete the table to the best of your
knowledge. Do not use more than one decimal place. If zero is one of
your answer, do not include any decimal. Enter only numbers or
signs. Graphical Help:
Additional help in completing the table can be obtained by clicking in
the bottoms of the applet on the left. By advancing the drawing
sequence, the corresponding time row of the table will be completed. 
Solution:
In order to calculate the velocity
of the ball, formula (F_F 3),
, should be used; in this case, the initial velocity is
.
At the same time, the displacement of the ball can be calculated from
formula (F_F 4),
.
After substituting the initial conditions in the equations, the
following formulas can be used for calculating the velocity and position
at the different times,
Velocity: 

Position: 


At time, t
= 0, 
{ 

The velocity and the displacement
are the initial conditions for the motion. 
At time, t
= 1, 
{ 

Since the ball is moving upward
and the acceleration of gravity is always downward, the ball is slowing
down. Because the displacement of the ball is above the zero level, the
displacement is positive. 
At time, t
= 2, 
{ 

At this time, the ball reaches
what is called the maximum height.
At this point in the motion the velocity of the ball is zero for an
instant. 
At time, t
= 3, 
{ 

Now, the ball is falling down;
thus, the velocity is negative. The magnitude of the velocity is
increasing because both the velocity and acceleration of gravity are
pointing in the same direction, down. 
At time, t
= 4, 
{ 

Here, the ball has returned to
the launching point (displacement zero meter), the edge of the building, the velocity has the same
magnitude as when launched. However, the sign is negative because the
direction of the velocity downward, still the ball is speeding up
because of the acceleration of gravity. In addition, at this point it
can be seeing that the equation of motion represents the displacement
not the distance traveled. In fact, at this point, the distance traveled
is 39.2 m; where 19.6 m of that distance is coming from the motion up
and 19.6 m coming from the motion down. 
At time, t
= 5, 
{ 

Finally, both, the velocity and
the displacement are negative. The displacement is negative because the
position of the ball is below the launching point. Notice that the ball
is still speeding up due to the acceleration of gravity. 
