ICSE Class 9 Answered

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 s

please give the explanation for this

Asked by pradeep.roy1959 | 03 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	$begin mathsize 14px style straight s equals 1 half cross times 10 cross times 0.6 squared equals 1.8 end style$