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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
7 years ago

Neville Daruwala
16 Points
```										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....
```
7 years ago
pratik nayak
33 Points
```										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

```
7 years ago
prakash pandey
33 Points
```										D ANSWER WILL B C OPTION USIND 2ND LAW..
```
7 years ago
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