Asked by Thakgalo sakwana
An object is projected vertically upwards and returns to the point of projection 12s later. Ignore air resistance and determine the velocity with which it was projected?
Answers
Answered by
Thakgalo sakwana
Vf=Vi +at
Vf=0, Vi=?, a=-9.8 remeber time taken to go up is equal to the time taken to go back to the point of projectile.. THEREFORE t=6
Vf=Vi +at
(0)=Vi +(-9.8)(6)
Vi=58.8m/s
***[OR]***
Vf=Vi +at
Vf=-Vi, Vi=?, a=-9.8, t=12
(-Vi)=Vi +(-9.8)(12)
-2Vi= -117.6
Vi=58.8m/s
if u've any question call 0723624826
Vf=0, Vi=?, a=-9.8 remeber time taken to go up is equal to the time taken to go back to the point of projectile.. THEREFORE t=6
Vf=Vi +at
(0)=Vi +(-9.8)(6)
Vi=58.8m/s
***[OR]***
Vf=Vi +at
Vf=-Vi, Vi=?, a=-9.8, t=12
(-Vi)=Vi +(-9.8)(12)
-2Vi= -117.6
Vi=58.8m/s
if u've any question call 0723624826
Answered by
Erick
It corect
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