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# with what speed should a body be thrown up so that the distance traversed in the 5th and 6th second is the same?

Sumit Majumdar IIT Delhi
7 years ago
Dear student,
Please confirm the direction along which the body is thrown. Has the body been projected vertically upwards or it moves horizontally or it has been projected at any angle with the horizontal. Depending upon the orientation of the angle of projection, we can find the distance covered.
Regards
Sumit
shashi K Sharma
46 Points
7 years ago
Dear Student

As per your question, you have that the body os projected up, taking that in consideration the body must be at maximum height at t=5second. As because from time height symmetry if the body is at same height at time= t1 and time= tthen (t1 +t2)= total time of flight.

5th second means between t = 4 & t= 5
6th second means  between t=5 & t=6
Distance covered in the two time gap stated above can be equal if and only if the body crosses maximum height position at t=5,
or the time of flight is 10 second or u=50m/s

ALTERNATE METHOD
Thus body is at same height at t=4 and t=6
Hence total time of flight is 10 second= 2u/g
thus u = 50m/s