The velocity at the maximum height of a projectile is half its initial velocity of projection v0. What is the horizontal range of the projectile?
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Answer:
√3u²/2g
Explanation:
Velocity at maximum height = half of initial velocity.
Velocity at maximum height = horizontal component.
v(X)=u/2
v(y)=√(u²-u²/4)
=√3u/2
v(y)/v(x)=√3
tanø=√3
ø=60°
R=2u²sinøcosø/g
=√3u²/2g
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