NEET Class neet Answered
A body is projected vertically up at t=0 with a velocity of 98m/sec. Another body is projected from the same point with the same velocity after time 4 sec. Both bodies will meet after :
Asked by Prashant DIGHE | 13 Jul, 2019, 10:14: AM
Expert Answer
Let after t seconds stating from projection time of first body, first body meets the second body.
Vertical displacement S is given by, S = u×t-(1/2)×g×t2 .....................(1)
where u is initial speed and g is acceleration due to gravity.
Vertical dispalcemnt for first body : S = 98×t -(1/2)×9.8×t2 = 98×t - 4.9×t2 ..................(2)
Vertical dispalcemnt for second body : S = 98×(t-4) -(1/2)×9.8×(t-4)2 = 98×(t-4) - 4.9×(t-4)2 ..................(3)
By equating (2) and (3), after simplification, we get , 392 = 4.9×[ t2 - (t-4)2 ] = 4.9×(2t-4)×4 ..............(4)
By solving eqn.(4), we get t = 12 s
Hence after 12 s, starting from projection time of first body, both the bodies will meet.
Answered by Thiyagarajan K | 13 Jul, 2019, 03:45: PM
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