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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/h18 meters from the ground please explain

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/h18 meters from the ground


 please explain

Grade:11

7 Answers

Neville Daruwala
16 Points
10 years ago

hey..according to ur question,it takes T time to travel a height H...therefore in T/3 sec it will travel a distance of h/9...using 2nd law of motion..i.e u=0,s=0+1/2gt^2...therefore its distance from ground will be H-H/9=8H/9....

pratik nayak
33 Points
10 years ago
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)
prakash pandey
33 Points
10 years ago

D ANSWER WILL B C OPTION USIND 2ND LAW..

Abhradeep
11 Points
2 years ago
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)
Shareef
15 Points
2 years ago
Accroding to 2 nd kinematic equation the body will travel 1/16 th of total height in t/4 seconds
=>this position of the body from the ground is = 
h-h/16=15/16h
Piyush Upadhyay
13 Points
2 years ago
As we know that ball is dropped from height H A and takes T seconds to fall on the ground, so it's initial velocity (u) will be 0.
So from 2 equation of motion we get, 
H=1/2gt2, as u=0........ (1)
Now let's consider the position of ballfrom air in T/3 seconds be H`. So again using 2 equation of motion, we get
H`=1/2gT2/9, as u=0.......(2)
Now from (1) and (2), we get
H`=H/9metres from the air.. 
So the location of ball from the ground will be=
H-H/9=9H-H/9=8H/9.
HENCE AFTER T/3 SECONDS POSITION OF BALL WILL BE 8H/9 METRES FROM THE GROUND.... 
 
Rishi Sharma
askIITians Faculty 646 Points
10 months ago
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

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