A shell is fired from point o at an angle of 60° with the horizontal with the speed 40 m/s
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Explanation:
ANSWER
During a projectile motion, horizontal velocity component is given by ucosθ.
The horizontal component of velocity u is ucosθ.
The horizontal component of velocity at the highest point is thus ucosθ.
Using momentum conservation,
2mu=mv
1
+mv
2
⇒2mucosθ=−mucosθ+mv
2
⇒mv
2
=2mucosθ+mucosθ=3mucosθ
⇒v
2
=
m
3mucosθ
=3ucosθ
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