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Question

# The equation of trajectory of a projectile is y=(x√3−gx220), where x and y are in meter. The maximum range of the projectile is (take g=10 m/s2)

A
83 m
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B
43 m
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C
34 m
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D
38 m
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Solution

## The correct option is B 43 mThe given equation of trajectory of projectile is y=(x√3−gx220)........(i) We know that the equation of trajectory is y=xtanθ−gx22u2cos2θ...........(ii) Comparing eq (i) & (ii) we get tanθ=1√3⇒ θ=30∘ Also, 2u2cos2θ=20 ⇒ u2=202cos230∘=403 m2/s2 As we know that the maximum range of the projectile is given by Rmax=u2g=40/310=43 m

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