Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
rajan jha Grade: 12

6 years ago

## Answers : (1)

vikas askiitian expert
510 Points

let its velocity  was v when it was in circular motion...after the string breaks , stone will move parabolically...

H = 2.7 (above the ground)

Ux= v       (velocity in circular path)

ay = -g

for horizontal motion threre is no accleration so ax =0

for verticle motion ,

S = gt2/2

2.7 = gt2/2

t = 0.74sec

now for horizontal motion

speed = distance/time

Ux = 10/(0.74)                                   (distance 10m (given))

Ux = 13.51m/s

with this velocity particle was circulating .....

ac(centripital accleration) = v2/r = (13.51)2/1.5 = 122 m/s2

6 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

## Other Related Questions on Mechanics

View all Questions »
• Complete Physics Course - Class 12
• OFFERED PRICE: Rs. 2,756
• View Details
• Complete Physics Course - Class 11
• OFFERED PRICE: Rs. 2,968
• View Details

## Ask Experts

Have any Question? Ask Experts

Post Question

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