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
lisa tulip Grade: Upto college level
`        A particle of mass 2m is projected at an angle 45 degree with the horizontal with a velocity of 2root2 m/s. after 1 s explosion takes place and the particle is borken into two equal pieces. as a result one part comes to rest. find the maximum height attained by the other part. `
8 years ago

Ramesh V
70 Points
```										The time of flight = 2u.sin x /g
=2*2.828*0.707 /10
=0.4 seconds
Making correction to 0.1 seconds, where explosion takes place
before explosion,
Vx =Vy = 2m/sec
before 0.1 sec,  Vx = 2 m/sec
Vy = 2 -0.1*10 = 1 m/sec
hz. distance covered = 2*0.1 = 0.2 m
vt.. distance covered = 2*0.1 - 5*0.01 = 0.15mts
before 0.1 sec,  explosion takes places
hz. and vt. moments shld be balanced which gives
in X dirn:    2m(2) = m(Vx' )
in Y dirn:     2m(1) = m(Vy' )
so, Vx' = 4 m/sec  and  Vy' =2 m/sec
now, max. height attained is hmax =   (Vy' )2 /2g
=4/20 = 0.2 m
so max. height attained from grnd level = 0.15 + 0.2 = 0.35 mts
---
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and
we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.

Regards,
Naga Ramesh
IIT Kgp - 2005 batch

```
8 years ago
Think You Can Provide A Better Answer ?

## 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