1
Question
A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball at T/3 second?
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Solution
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 +1/2gT2
h = 1/2gT2 ------[1]
in T/3 sec h1 = 1/2gT2/9 -------[2]
from [1] and [2] we get h1 =h/9 distance from point of release.
therefore distance from ground is h-h/9 =8h/9
in T sec. body travels h mts.
by applying equations of motion we get
s= ut +1/2gT2
h = 1/2gT2 ------[1]
in T/3 sec h1 = 1/2gT2/9 -------[2]
from [1] and [2] we get h1 =h/9 distance from point of release.
therefore distance from ground is h-h/9 =8h/9
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