A ball thrown up vertically returns to the thrower after 6 s. Find. its position after 4 s.
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Answered by
2
SEE
IF THE BALL IS THROWN VERTICALLY...
IT REACHES THE GROUND IN 6S....
THE TIME FOR GOING UP AND REACHING THE GROUND IS 3S ......
SO IN THE 1ST 3 S IT REACHED THE MAX. HEIGHT AND AFTER 1S.....IT WILL IN THE AIR .....
IN THE NON SURFACE AREA ......
SO THE POSITION OF BALL WILL BE IN THE AIR .......☺
IF THE BALL IS THROWN VERTICALLY...
IT REACHES THE GROUND IN 6S....
THE TIME FOR GOING UP AND REACHING THE GROUND IS 3S ......
SO IN THE 1ST 3 S IT REACHED THE MAX. HEIGHT AND AFTER 1S.....IT WILL IN THE AIR .....
IN THE NON SURFACE AREA ......
SO THE POSITION OF BALL WILL BE IN THE AIR .......☺
Answered by
5
Position of the ball after 4s is From Second equation of motion s = ut + 1/2 gt2 or, s = (29.4X4) + 1/2 X (-9.8)(4)2 or, s = 117.6 - 78.4 = 39.2 m
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