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Consider an object thrown upwards from the ground with the initial velocity of \( 30\frac{\mathrm{m}}{\mathrm{s}} \). The object moves upwards with decreasing vertical velocity until it stops. Then it starts moving vertically downwards. Find the greatest height above the ground the object does reach. \[ \] Note: The vertical distance \( y \) of a thrown object is described by the equation \( y=v_0t-\frac12gt^2 \), where \( v_0 \) is the initial velocity of the thrown object, \( g \) is gravitational acceleration (count with the rounded value \( 10\frac{\mathrm{m}}{\mathrm{s}^2}\)), and \( t \) is the time period of the object motion in seconds.