(for a better understanding of engineering dynamics problems ,under graduate coarse)

1. ABSTRACT:
Studying the kinematics of a simple problem as stated above. I will be working with free fall
motion in the above example.
ANALYSIS:
 The ball is thrown upward with
1. Vi=15
푚
푠
2. a=-9.8
푚
푠2
3. Vf=0
For constant acceleration problems
Vf=Vi+at , t=
15
9.8
=>1.5306 @maximum height
S=Vit-0.5a푡2 , s=11.479 m
 Now here we have the deceleration problem now If it were a positive acceleration we
would have more distance traveled as initial velocity was given since then it would have
larger range in certain domain but for deceleration problem the range is restricted to
the maximum distance object would cover with zero acceleration and 15
풎
풔
of initial
velocity for 1.5306 sec
15
10
5
0
s-t
0 0.5 1 1.5 2
distance meters
time sec
Series1

2. 20
15
10
5
0
v-t
0 0.5 1 1.5 2
velocity m/s
SECOND PHASE OF MOTION:
time sec
Series1
Now if the object is not catched at the height of 11.479 meters the object will fall as free falling
object and for that we have generalized s-t and v-t graphs as:
600
400
200
0
-200
distance
0 2 4 6 8 10 12
time sec
distance
Poly. (distance)
y = 9.8x - 2E-13
150
100
50
0
-50
velocity
0 2 4 6 8 10 12
time sec
velocity
Poly. (velocity)
That is for free fall the object have to touch the gorund to be stopped ,in that case it may be the
distance meters
velocity meter/sec
distance of 11.47 meters or if there is a case that this height was just achieved on the top of multi story
building then the object will continue to fall further down.

3. ANNEXURE:
TABLE 1:
t s Vi a
0 0 0 9.8
0.255 0.318623 2.499 9.8
0.51 1.27449 4.998 9.8
0.765 2.867603 7.497 9.8
1.02 5.09796 9.996 9.8
1.275 7.965563 12.495 9.8
1.53 11.47041 14.994 9.8