Instantaneous results!
1. a.
A ball is projected vertically upload with velocity19.6 m/s. it
goes to 19.6 m high and come to ground in 4 sec. Find the velocity of ball at 1
sec..
b.
In this experiment a ball is thrown from rest to reach a certain
height and at a certain velocity before starting to come down.
c.
The table is look like
|
TIME
|
HEIGHT
|
|
0
1
2
3
4
|
0
9.8
19.6
14.7
0
|
The table is look like
TIME
|
HEIGHT
|
0
1
2
3
4
|
0
9.8
19.6
14.7
0
|
d. The graph of the table will be like…..
e.
In the above graph the three secant line is can be obtain from
these four points
Let
points A (1, 9.8), B (2, 19.6) C (3, 14.7) and D (4, 0)
For the line AB the slope will
be
M1 = (19.6 – 9.8)/(2 - 1)
= 9.8
For line AC the
slope will be
M2 = (14.7-9.8)/(3 - 1)
= 2.45
For line AD the slope
will be
M3 = (4 – 9.8)/(4 - 1)
= - 1.9333
Therefore the graph with the secant line will be shown as:
To find instantaneous rate of change of height with respect to
time can be calculated by taking another point Q on the secant line and find
the slope of line joining point A and Q. the point A is one through which all the secant line
has passed.
f.
This calculation that is
calculation of rate of change of height with respect to time will be the
velocity of the particle. The IRC or derivative at a point on the given graph will
give the instantaneous velocity of the particle at given point.
In the calculation of IRC or derivative at a
point we have taken height a numerator
and time as denominator. Hence the unit of IRC or derivative at a point will be
meter/second.
g.
From the calculation we have seen that the
rate of change of height with respect to time between t = 0 to t= 1 is 9.8 m/s
. this shows the average velocity between these to time instant.
h.
Similarly we have rate of change of
height with respect to time for all interval. Hence we have average velocity
throughout the motion.
Therefore velocity of the particle at time t = 1 sec will be 9.8 m/s.
Very good, I like how you used the graphs to your advantage. I wish i had the resources you did, it would have made my work much easier. I did a similar post on rocket trajectory but the data i got was different than yours.
ReplyDeleteOverall really good job, but I was slightly confused by your graph with the tangent line. I wasn't sure if it was there or if you maybe left it out. Good job though!
ReplyDeleteit is a great work and I found it very interesting.
ReplyDeletemlak,
ReplyDeleteyour application is a good one and your tables are well done. i couldn't see some of your calculations in part f, but it appears that they are correct, given the values that you explained at the end of your post. it would have been good to explain what the secant line calculations are telling you and how they compare to the IRC calculation, but other than that, good job.
professor little