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.

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,

V_x =V_y = 2m/sec

before 0.1 sec,  V_x = 2 m/sec

                         V_y = 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(V_x' )

in Y dirn:     2m(1) = m(V_y' )

so, V_x' = 4 m/sec  and V_y' =2 m/sec

now, max. height attained is h_max = (V_y' )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