General Physics> kinematics...

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

aditya bhardwaj , 14 Years ago

Grade 11

7 Answers

Neville Daruwala

Last Activity: 14 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

Last Activity: 14 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)

Approved

prakash pandey

Last Activity: 14 Years ago

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

Abhradeep

Last Activity: 6 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

Last Activity: 6 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

Last Activity: 6 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

Last Activity: 4 Years 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