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

### Mohammed

Guys, does anyone know the answer?

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

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

Suggest Corrections 64

## 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 beQuestion

(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=2HBut 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 .

Was this answer helpful?

205 32

## 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)

Home > English > Class 12 > Physics > Chapter > Cengage Physics Dpp >

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

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

## Related Videos

11296667 0 800 3:15

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

645862818 0 8.7 K 2:55

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

11745877 0 2.6 K 6:17

A particle is projeted 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 :

15792421 0 9.1 K 2:04

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)

647778913 0 1.1 K

A ball is projected with a velocity v so that its range on a horizontal plane is twice the greatest height attained. Then, its range is :

409497097 0 8.8 K 3:52

A particle is projected with a velocity v so that its horizontal range twice the greatest height attained. The horizontal range is

Show More Comments

Add a public comment...

Follow Us:

Popular Chapters by Class:

Class 6 Algebra

Basic Geometrical Ideas

Data Handling Decimals Fractions Class 7

Algebraic Expressions

Comparing Quantities

Congruence of Triangles

Data Handling

Exponents and Powers

Class 8

Algebraic Expressions and Identities

Comparing Quantities

Cubes and Cube Roots

Data Handling

Direct and Inverse Proportions

Class 9

Areas of Parallelograms and Triangles

Circles Coordinate Geometry Herons Formula

Introduction to Euclids Geometry

Class 10

Areas Related to Circles

Arithmetic Progressions

Circles Coordinate Geometry

Introduction to Trigonometry

Class 11 Binomial Theorem

Complex Numbers and Quadratic Equations

Conic Sections

Introduction to Three Dimensional Geometry

Limits and Derivatives

Class 12

Application of Derivatives

Application of Integrals

Continuity and Differentiability

Determinants

Differential Equations

Privacy Policy

Terms And Conditions

Disclosure Policy Contact Us

Guys, does anyone know the answer?