A shell is fired from a cannon with a speed of `100 m//s` at an angle `60^(@)` with the horizontal (positive `x`-direction). At the highest point of its trajectory, the shell explodes in to two equal fragments. One of the fragments moves is the speed of the other fragment at the time of explosion.

A shell is fired from a cannon with a speed of `100 m//s` at an angle

1.	A shell is fired from a cannon with a speed of `100 m//s` at an angle `60^(@)` with the horizontal (positive `x`-direction). At the highest point of its trajectory, the shell explodes in to two equal fragments. One of the fragments moves is the speed of the other fragment at the time of explosion.
Answer» The velocity of the shell at the highest point of tarajector is `v_(M)=ucostheta=100 cos 60^(@)=50 m//s` Let `v_(1)` be the speed of the fragment which moves long the negative `x`-direction and the other fragment has speed `v_(2)`, which must along the positive `x`-direction. now from, momentum conservation, we have `mv=(-m)/2 v_(1)=m/2v_(2)` or `2v=v_(2)-v_(1)` or `v_(2)=2v+v_(1)=(2xx50)+50=150m//s`

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A shell is fired from a cannon with a speed of `100 m//s` at an angle `60^(@)` with the horizontal (positive `x`-direction). At the highest point of its trajectory, the shell explodes in to two equal fragments. One of the fragments moves is the speed of the other fragment at the time of explosion.

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