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An observer sees a flower pot sail up and then back down past a window 2.45 m high. If the total time the pot is in the sight is 1 s, what is the height above the window upto which the pot rises ?
Dear Fawz naim ,
THe time the pot takes to go up and down through the window must be same as the conditions in both cases are same. The magnitude of velocity at the entry and exit points are same and distance travelled is same too . so time taken to go up and down = 0.5 s
now we can take the axis of reference anywhere on the trajectory of the motion , we take it on the bottom of window. Assuming the velociyt at that point to be u , and applying the equaitons of motion ,
2.45 = u*(.5) -(1/2) *9.8*.52
solving we get u = 7.35m/s
so the maximum height from the bottom of the window = u2/2g = 2.7625
height from the top of window is 2.7625- 2.45 = .306 m
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