NEET Class neet Answered
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Asked by Prashant DIGHE | 13 Jul, 2019, 10:04: AM
Expert Answer
When the lift is ascending with accelration a, the acceleration acting on the ball thrown vertically is (g+a).
When the lift is descending with accelration a, the acceleration acting on the ball thrown vertically is (g-a).
since the ball is caught by same person , displacement is zero.
Hence when lift is ascending, we have, ut1 - (1/2)(g+a)t12 = 0 or u = (1/2)(g+a)t1 ........................ (1)
when lift is descending, we have, ut2 - (1/2)(g+a)t22 = 0 or u = (1/2)(g-a)t2 ........................ (2)
hence from (1) and (2) , (g+a)t1 = (g-a)t2 or acceleration a = [ (t2 - t1) / (t1 + t2) ]×g
If we add eqn.(1) and (2), we get, 4u = (g+a)t1 + (g-a)t2 = g(t1 + t2) - a (t2 - t1) = g(t1 + t2) - [ (t2 - t1)2 / (t1 + t2 )] ..............(3)
after simplification, we get from eqn.(3), velocity u = g t2 t1 /( t1 + t2 ) .............................(4)
Answered by Thiyagarajan K | 13 Jul, 2019, 02:01: PM
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