A shell is shot with an initial velocity v of 20 m/s, at an angle of = 60 degree

Question

A shell is shot with an initial velocity v of 20 m/s, at
an angle of
= 60 degrees with the horizontal. At the top
of the trajectory, the shell explodes into two fragments of
equal mass (See Figure). 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 air drag is
negligible?

please include a free body diagram if possible . thanks

Explanation / Answer

momentum along horizontal is conserved .so initial along x = 20cos60=10m/s Momentum initial along x = m*v=m/2*v1 => v1=2v =20 At top distance horizontal is 10*10/10=10m At top v=20 horizontal distance is 20*10/10=20 total = 30m

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