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Question a ball thrown up vertically returns to the thrower after 6 seconds find the velocity with which it was thrown up the maximum height it reaches and the position |
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Answer» after analysis of questionu=?v=0t=6/2=3 seca=-9.8m/s2 therefore v=u+at0=u+(-9.8)(3)u=29.4m/sthe VELOCITY with which it was THROWN =29.4 m/s the DISTANCE it travelleds=ut + a(tsquare)/2 =29.4*3+(-9.8)(9)/2 =88.9+44.1 =187.3 m |
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