A ball is released from the top of a tower of height h meters. It takes T second to reach the ground. What is the position of the ball in T/3 second -
Answers
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the acceleration of the ball will be g. Initial velocity will be 0.
in T sec. body travels h mts.
by applying equations of motion we get
s= ut+ (1/2)gT
2
h= (1/2)gT
2
------[1]
in T/3 sec
h
1
(1/2)gT
2
=
(1/2)g(
3
T
) 2
(1/2)g(
9
T2
)
-------[2]
from
[1]and[2]we get
h
1
9
h
distance from point of release.
therefore distance from ground is
h−
9
h
=
9
8h
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Answer:
from [1] and [2] we get h1 =h/9 distance from point of release. therefore distance from ground is h-h/9 =8h/9. , studies Bachelor of Technology in Computer Science and Engineering at Maharaja Agrasen Institute of Technolog… Originally Answered: A ball released from top of the tower of height h metres .the acceleration of the ball will be g. Initial velocity will be 0.
in T sec. body travels h mts.
by applying equations of motion we get
s=ut+(1/2)gT
2
h=(1/2)gT
2
------[1]
in T/3 sec
h
1
=(1/2)gT
2
=(1/2)g(
3
T
)
2
=(1/2)g(
9
T
2
) -------[2]
from
[1] and [2] we get h
1
=
9
h
distance from point of release.
therefore distance from ground is h−
9
h
=
9
8h