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 f the ball in T/3 second –

(A) h/9 meters from the ground

(B) 7h/9 meters from the ground

(C) 8h/9 meters from the ground

(D) 17/div8 meters from the ground

please explain

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
the answer is (C)

initial velocity (u)=0Acceleration=acceleration due to gravity(g)Time=T So by the equation of motionH=ut+1/2gT^2 so H=1/2 gT^2.......[1]And distance or H at T/3 seconds isH=u×T/3+ 1/2×g×T^2/9So by solving H/9=1/2gT^2......,[2]So,distance covered at T/3 seconds is H/9And from the point of release is equation [1]-[2] which isH-H/9=(9H-H)/9=8H/9Hence,the answer is option (C)

Dear Student,
Please find below the solution to your problem.

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
h1​ = (1/2)gT^2
= (1/2)g(T/3​)^2
= (1/2)g(T^2​)/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

Thanks and Regards