A shell is fired from a gun with a muzzle velocity of 14 m/s, at an angle of 60° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass (Fig. 9-30). One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that the air drag is negligible?

We will use momentum conservation to solve this problem.
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<br>Since m1 falls straight down from the position at which the shell reaches its maximum height, we need to find the horizontal position when Vy = 0: (no vertical motion)
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<br>Vy(t) = Vo(in y direction) - g t
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<br>Or, 0 = Vo Sin(theta) - g t
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<br>Or, t = Vo*Sin(theta)/g -----------------(1)
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<br>Now, the horizontal position at this time is given by (taking xo to be 0):
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<br>x(t) = Vo(x-component,t) * t = ...

Solution Summary

Not just an answer. Explanations are provided, with a figure.

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