A stone is thrown vertically upward with an initial velocity v0.The distance travelled in time 4v0/3g is
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Hey mate here is ur solution...
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As the acceleration due to gravity is -g
hence travelled distance can be found by formula s = ut +½at²
travelled height in 4v0/3g sec is
H= v0*4v0/3g -½g *(4v0/3g )²
=4v0²/3g-½g* 16v0²/9g²
= 4v0²/3g -8*v0²/9g
=4v0²/9g this is travelled distance in given time
_________________
Hope this will help you...
________________
As the acceleration due to gravity is -g
hence travelled distance can be found by formula s = ut +½at²
travelled height in 4v0/3g sec is
H= v0*4v0/3g -½g *(4v0/3g )²
=4v0²/3g-½g* 16v0²/9g²
= 4v0²/3g -8*v0²/9g
=4v0²/9g this is travelled distance in given time
_________________
Hope this will help you...
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