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    a particle is projected with a velocity v such that its range on the horizontal plane is twice

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    Question

    A particle is projected with a velocity

    v

    so that its range on a horizontal plane is twice the greatest height attained. Consider

    g = 10 m/s 2

    as acceleration due to gravity. Then, its range will be:

    Video Solution Open in App Solution

    The correct option is A

    4 v 2 5 g Range is given by R = u 2 sin 2 θ g

    and maximum height is given by

    H = u 2 sin 2 θ 2 g and R = 2 H [Given in question] ⇒ u 2 sin 2 θ g = 2 × u 2 sin 2 θ 2 g ⇒ 2 sin θ cos θ = sin 2 θ ⇒ tan θ = 2

    From triangle A B C , sin θ = 2 √ 5 , cos θ = 1 √ 5 ∴ R = 2 u 2 sin θ cos θ g = 2 v 2 × 2 √ 5 × 1 √ 5 g ⇒ R = 4 v 2 5 g

    Projectile Time, Height and Range

    Standard XII Physics

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    स्रोत : byjus.com

    A particle is projected with a velocity u 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).

    Click here👆to get an answer to your question ✍️ A particle is projected with a velocity u 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 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

    Question

    (g is the acceleration due to gravity).

    A

    5g 4v 2 ​

    B

    5g ​ 4v 2 ​

    C

    g v 2 ​

    D

    g v ​ ​ Medium Open in App

    Updated on : 2022-09-05

    Solution Verified by Toppr

    Correct option is A)

    Given, Range, R=2H

    But R=4Hcotθ ⇒cotθ= 2 1 ​

    From triangle we can say that,

    sinθ= 5 ​ 2 ​ and cosθ= 5 ​ 1 ​

    So, the range of the projectile

    R= g 2v 2 sinθcosθ ​ = g 2v 2 ​ × 5 ​ 2 ​ × 5 ​ 1 ​ = 5g 4v 2 ​ .

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    स्रोत : www.toppr.com

    A particle is projected with a velocity v so that its range on a horizontal plane is twice the greatest height attained. If g is acceleration due to gravity, then its range is

    R=2H given We know R=4Hcotthetaimpliescottheta=(1)/(2) From triangle we can say that sintheta=(2)/(sqrt5),costheta=(1)/(sqrt5) Range of projectile R=(2v^2sinthetacostheta)/(g) =(2v^2)/(g)xx(2)/(sqrt5)xx(1)/(sqrt5)=(4v^2)/(5g)

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    A particle is projected with a...

    A particle is projected with a velocity

    v v

    so that its range on a horizontal plane is twice the greatest height attained. If

    g g

    is acceleration due to gravity, then its range is

    Updated On: 27-06-2022

    00 : 30

    UPLOAD PHOTO AND GET THE ANSWER NOW!

    Text Solution Open Answer in App A 4 v 2 5g 4v25g B 4g 5 v 2 4g5v2 C v 2 g v2g D 4 v 2 5 – √ g 4v25g Answer

    The correct Answer is A

    Solution R=2H R=2H given We know R=4Hcotθ⇒cotθ= 1 2 R=4Hcotθ⇒cotθ=12

    From triangle we can say that

    sinθ= 2 5 – √ sinθ=25 , cosθ= 1 5 – √ cosθ=15 Range of projectile R= 2 v 2 sinθcosθ g R=2v2sinθcosθg = 2 v 2 g × 2 5 – √ × 1 5 – √ = 4 v 2 5g =2v2g×25×15=4v25g

    Answer

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