Question

a shell is fired from a Cannon at an angle of 30 degree with horizontal velocity of 1000 metre per second assuming G equal to 10 metre per second find time taken by the shell to reach the highest point and how high the shell will arise..​

Answer 1

Explanation:

Solution :

The velocity of shell at the highest point is v=usinθ=100×sin30∘=50m/sv=usinθ=100×sin30∘=50m/s. Let in be the mass of the shell. Then the mass of the lighter fragment is m/3m/3 and that of heavier fragment is 2m/32m/3. Initial momentum of the shell before explosion is

mv=50mmv=50m

As no external force is acting on the shell, we can conserve momentum of the shell before and after its explosion.

In xx-direction

mv=2m3v2cosθ⇒v2cosθ=32vmv=2m3v2cosθ⇒v2cosθ=32v..............i

in yy- direction

0=m3v1−2m3v2sinθ⇒v2sinθ=v120=m3v1-2m3v2sinθ⇒v2sinθ=v12........ii ltbr Squar∈gandadd∈geqniandiiwe≥tSquar∈gandadd∈geqniandiiwe≥tv_(2)^(2)=9/4v^(2)+1/4v_(1)^(2)implies v_(2)=1/2sqrt(9v^(2)+v_(1)^(2))=1/2sqrt(9xx2500

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