A covering is shot with an early stage velocity 0 that 20 m/s, at an edge of θ0 = 60° with the horizontal. In ~ the peak of the trajectory, the covering explodes right into two fragments of same mass . One fragment,whose speed immediately after the to explode is zero, drops vertically. How much from the gun does the various other fragment land, assuming that the terrain is level andthat air drag is negligible?

The ideas used to deal with this difficulty are projectile motion and the law of preservation of momentum.

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Initially, usage the expression in terms of initial velocity and the edge made by the early stage velocity with the horizontal to determine the horizontal component of the early stage velocity the the shell.

Finally, usage the regulation of conservation of momentum to recognize the speed of 2nd fragment instantly after the explosion.

The horizontal ingredient of the early stage velocity in a projectile activity is offered as follows: Here, the horizontal ingredient of the early velocity is , the initial velocity is , and also the angle made through the early stage velocity through the horizontal is .

According come the law of conservation of momentum, “the total momentum of one isolated system remains the same”.

The inert of a human body is offered by the expression together follows: Here, the inert is , mass of the human body is , and its velocity is .

The horizontal ingredient of the early stage velocity in a projectile movement is given as follows: Substitute because that and also because that . According to the legislation of preservation of momentum, “the total momentum of an isolated mechanism remains the same”.

The initial momentum of the covering at the top of the trajectory prior to explosion is provided as follows: Here, the initial momentum is .

The shell explodes right into two pieces of same mass at the peak of the trajectory.

The masses that the two fragments will be half of the fixed of the shell.

Thus: Here, the fixed of the first and second fragments room and , respectively.

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The last momentum at the height of the trajectory ~ explosion is given as follows: Here, the final momentum is , the velocity the the an initial and 2nd fragments are and , respectively.