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`        A ball of mass 2kg, charge 1×10-6 C is dropped from top of a high tower . In space electric field exists in horizontal direction away from the tower whoch vareis as E=(5-2x)×106 V/m. Find the maximum horizontal distance ball can go from the tower and displacement of ball at this instant from the point of projection?`
8 years ago

Pratham Ashish
17 Points
```										hi ,
look the figure ,below . it shows only how the horizonta motion occurs
dv/dt = (5-2x)×106   ( 1×10-6   )
=(5-2x)
dx/dt = v
dv/dx = (5-2x)/v
v dv = (5-2x) dx
v^2/ 2]  from 0 to v =  5x - x^2 ]  from o to x
v^2/ 2  =  5x - x^2
for v =0 ,
5x - x^2 = 0
x= 0,  5
so ball will move back at x= 5 , this will be max. horizontal distance
displacement of ball at this instant from the point of projection ,
dx/ dt = v = √ 10x -  2 x^2
from here we can get time when it reaches  x= 5 ,
in this duration vertical displacement =  1/2 g t ^2
total  displacement  =  √ ( horiz. displacemen ) ^2 + ( vertical displacement  ) ^2

```
8 years ago
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