A Particle projected vertically upward passes the same height h at 2s and at 10s.If the acceleration due to gravity g=9.8 m/s^2, then what is the height?

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Solution

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 flight

time of upward journey =(t1+t2)/2

now v=u+at

v at top most point is 0 and a is negative

therefore, u=at

or u=9.8 X 2=58.8

now H =ut+1/2at^2

puttting t=2 and u=58.8 and a=-9.8 we get,

H=98 metres.

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