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Question

# An object is allowed to fall freely from a tower of height 39.2m; exactly at the same time another stone is thrown from the bottom of the tower in vertically upward direction with a velocity of 19.6m s−1 Calculate when and where these two stones would meet ?

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Solution

## Given that,Height h=39.2m Velocity u=19.6m/sNow, from equation of motionCase (I) when stone dropped from the tower fall freely s1=ut+12gt2 s1=0+12gt2....(I) Now, case (II) when stone projected upwards1=19.6t−12gt2....(II) Now, total distance s=s1+s2 s=12gt2+19.6t−12gt2 39.2=19.6t t=2sec Now, put the value of t in equation (I) s1=12×9.8×4 s1=19.6m Now, the distance is s=s1+s2 s2=39.2−19.6 s2=19.6m Hence, Both stones meet from ground in air at the distance of 19.6 m in 2 sec

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