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Distance travelled in t second,
Here, initial velocity u=0h=gt22 ...... (1)Distance travelled in nth second = u +a2(2n -1)Distance travelled in last second= 0 + g2(2t-1) =9h25 ..............(2)Equating h from equation 1 and 225g(2t-1)18=gt229t2 -50t +25 =0Solving for t,t =50 ±502-4×9×252×9 = 1018s or 5 sTime has to be greater than 1 second, so t=5sPutting value of t in eqn. 1h=g522 = 122.5 m