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A particle is projected vertically upward passes the same same height at 2sec and at 10 sec if acceleration due to gravity is 9.8met per sec2 then height is equal to
The motion under gravity is a symmetric motion therefore particle will take same time in covering same distance in upward and downward journey.therefore if it is at a height H at first 2 sec, in upward journey, it will be at the same height H in the last 2 sec of downward jouney.therefore, t1+t2= total time of flighttime of upward journey =(t1+t2)/2now v=u+atv at top most point is 0 and a is negative therefore, u=ator u=9.8 X 2=58.8now H =ut+1/2at^2puttting t=2 and u=58.8 and a=-9.8 we get,H=98 metres.
Height covered by particle in 2sec = height covered in 10secSo,. By S= u.t + 1/2 (a.t)^2u.2 + 1/2 (-9.8) (2×2) =. u.10 +1/2 (-9.8) (10×10)2u - 19.6 =. 10u - 49010u-2u. =490-19.68u. = 470.4u. =. 58.8 m/secBy putting value of `u` in above formula2×58.8. -1/2 ×9.8×4117.6 -19.698 m
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