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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 ​ .

205 32

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

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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

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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