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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

6 years ago


Answers : (3)


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....

6 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                      h= 1/2gT2/9                                     -------[2]

from [1] and [2] we get h=h/9 distance from point of release.

therefore distance from ground is h-h/9 =8h/9

the answer is (C)


6 years ago


6 years ago

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