Question

A particle of mass 2m is projected at an angle of with the horizontal with a velocity of . After 1s of explosion, the particle breaks into two equal pieces. As a result of this one part comes to rest. The maximum height from the ground attained by the other part is​

Answer 1

Answer:

Answer

put formula velocity and velocity answer divide by mass formula answer should be Understand

Explanation:

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Answer 2

Answer:

35  m

Explanation:

Given :  v=202  ms−1

 ∴ Vertical component of velocity  u=vsin45o=20 ms−1

Let the height attained by  2m in 1 sec be  h1

∴    h1=ut−21gt2

h1=20(1)−21(10)(1)2            

⟹h1=15 m

Now at point A, this breaks into two equal pieces such that one comes to rest and the other moves upward with velocity v′

Applying conservation of momentum in y direction :

(2m)u=0+m(v′)             

v′=2u=2(20)=40   ms−1

Let the height above point A that the piece attains be  h′

∴    0−(v′)2=2(−g)h′

−402=2(−10)h′                ⟹h′=20  m

Thus total height attained by the piece       H=h1+h′=15+20=35  m