MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
Srijit singh Grade: 12
        

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?

7 years ago

Answers : (1)

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


 


 


4604-741_4788_19.bmp

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: R 2,600
  • View Details
  • Complete Physics Course - Class 11
  • OFFERED PRICE: R 2,800
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details