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

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

Answers : (1)

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.


Naga Ramesh

IIT Kgp - 2005 batch


8 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • 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
To Win!!! Click Here for details