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Question
A particle is projected with a velocity v such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g is acceleration due to gravity)
A particle is projected with a velocity v such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g is acceleration due to gravity)
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Solution
The correct option is A
R = 2H given
We know R = 4H cot θ ⇒ cot θ = 12
From triangle we can say that sin θ = 2√5, cos θ = 1√5
∴ Range of projectile R = 2v2sinθcosθg
= 2v2g × 2√5 × 1√5 = 4v25g
R = 2H given
We know R = 4H cot θ ⇒ cot θ = 12
From triangle we can say that sin θ = 2√5, cos θ = 1√5
∴ Range of projectile R = 2v2sinθcosθg
= 2v2g × 2√5 × 1√5 = 4v25g
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