A particle is projected with a velocity v such that its range on horizontal plane is twice the greatest height attained by it. The range of the projectile will be (g is the acceleration due to gravity).
A
4v25g
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B
4g5v2
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C
v2g
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D
4v2√5g
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Solution
The correct option is A4v25g From the given condition of the projectile we have R=2H
Where, R is the range of the projectile and H is the height of the projectile.
We know that, R=4Hcotθ⇒cotθ=2H4H=12
Plotting the value of cotθ on triangle we get
sinθ=2√5,cosθ=1√5 ∴ Range of projectile is R=2v2sinθcosθg =2v2g×2√5×1√5=4v25g.