a body at rest is dropped from the top of a tower .draw a displacement -time graph of its free fall under gravity during the first 0.6 seconds.show table for displacement and time starting from t=0 with an interval of 0.1 second.take g=10 m sblank to the power of negative 2 end exponent
please give the explanation for this

Asked by pradeep.roy1959 | 3rd May, 2015, 08:20: PM

Expert Answer:

The body is dropped from rest. Hence, u = 0
 
When the body is dropped, its speed increases and so does its displacement.
 
From the third equation of motion, we have begin mathsize 14px style straight s equals ut plus 1 half at squared equals 0 plus 1 half at squared equals 1 half at squared end style
The table for first six seconds of the fall of the body is shown below.

 Time (t in seconds)

Displacement (s in metres) 

 0

begin mathsize 14px style straight s equals 1 half cross times 10 cross times 0 squared equals 0 end style

 0.1

 begin mathsize 14px style straight s equals 1 half cross times 10 cross times 0.1 squared equals 0.05 end style

 0.2

begin mathsize 14px style straight s equals 1 half cross times 10 cross times 0.2 squared equals 0.2 end style

 0.3

 begin mathsize 14px style straight s equals 1 half cross times 10 cross times 0.3 squared equals 0.45 end style

 0.4

 begin mathsize 14px style straight s equals 1 half cross times 10 cross times 0.4 squared equals 0.8 end style

 0.5

 begin mathsize 14px style straight s equals 1 half cross times 10 cross times 0.5 squared equals 1.25 end style

 0.6

 begin mathsize 14px style straight s equals 1 half cross times 10 cross times 0.6 squared equals 1.8 end style

 
The displacement-time graph for the motion of the body is as shown below.

 
The graph will be a curve starting from origin as displacement is directly proportional to square of time.

Answered by Romal Bhansali | 4th May, 2015, 12:28: PM