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PHXI04:MOTION IN A PLANE

362017 Assertion :
When $θ = 45^{\circ}$ or $135^{\circ}$ , the value of $R$ remains the same, only the sign changes.
Reason :
$R = \frac{u^{2} \sin 2 θ}{g}$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Horizontal range, $R = \frac{u^{2} \sin 2 θ}{g}$
When $θ = 45^{\circ}$ , maximum horizontal range,
$R_{max} = \frac{u^{2}}{g} \sin 90^{\circ} = \frac{u^{2}}{g}$
When $θ = 135^{\circ}$ , maximum horizontal range,
$R = \frac{u^{2}}{g} \sin 270^{\circ} = \frac{- u^{2}}{g}$
(Negative sign implies opposite direction)

AIIMS - 2017

PHXI04:MOTION IN A PLANE

362018 For an object thrown at $45^{\circ}$ to horizontal, the maximum height ( $H$ ) and horizontal range ( $R$ ) are related as

1

R = 16 H

2

R = 8 H

3

R = 4 H

4

R = 2 H

Explanation:

For $θ = 45^{\circ}$
$H_{max} = \frac{u^{2} \sin^{2} 45^{\circ}}{2 g} = \frac{u^{2}}{4 g}$ $[∵ \sin 45^{\circ} = \frac{1}{\sqrt{2}}]$
$R = \frac{u^{2} \sin 90^{\circ}}{g} = \frac{u^{2}}{g}; ∴ \frac{R}{H} = \frac{u^{2}}{g} \times \frac{4 g}{u^{2}} = 4 \Rightarrow R = 4 H$

PHXI04:MOTION IN A PLANE

362019 Statement A :
When the range of projectile is maximum, the time of flight is the largest.
Statement B :
Range is maximum when angle of projection is $45^{\circ}$

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

$T = \frac{2 u \sin θ}{g}$ , it will maximum, when $θ = 0^{\circ}$ .
$R_{max} = \frac{u^{2}}{g}, f o r θ = 45^{\circ}$
So option (2) is correct.

PHXI04:MOTION IN A PLANE

362020 The ceiling of a long hall is 25 $m$ high. Then, the maximum horizontal distance that a ball thrown with a speed of 40 $m$ / $s$ can go without hitting the ceiling of the hall, is

1

95.5 m

2

105.5 m

3

100 m

4

150.5 m

Explanation:

Given, initial velocity $(u) = 40 m / s$
Height of the hall $(H) = 25 m$
Let the angle of projection of the ball be $θ$ , when maximum height attained by it be 25 $m$ .
Maximum height attained by the ball
$H = \frac{u^{2} \sin^{2} θ}{2 g}$
$25 = \frac{{(40)}^{2} \sin^{2} θ}{2 \times 9.8}$
or $\sin^{2} θ = \frac{25 \times 2 \times 9.8}{1600}$
$= 0.3063$
or $\sin θ = 0.5534$
$= \sin {33.6}^{\circ}$
or $θ = {33.6}^{\circ}$
$∴$ Horizontal range $(R) = \frac{u^{2} \sin 2 θ}{g}$
$= \frac{{(40)}^{2} \sin 2 \times {33.6}^{\circ}}{9.8}$
$= \frac{1600 \times \sin {67.2}^{\circ}}{9.8}$
$= \frac{1600 \times 0.9219}{9.8} = 150.5 m$

NCERT Exemplar

PHXI04:MOTION IN A PLANE

362021 The range of the projectile projected at an angle of $15^{\circ}$ with horizontal is $50 m$ . If the projectile is projected with same velocity at an angle of $45^{\circ}$ with horizontal, then its range will be

1

100 \sqrt{2} m

2

100 m

3

50 \sqrt{2} m

4

50 m

Explanation:

$R = \frac{u^{2}}{g} \sin 2 θ$
$∴ \frac{R_{45}}{R_{15}} = \frac{\sin 2 \times 45^{\circ}}{\sin 2 \times 15^{\circ}}$
$\Rightarrow R_{45} = 50 \times \frac{1}{1 / 2} m$
$= 100 m$

JEE - 2023

PHXI04:MOTION IN A PLANE

362017 Assertion :
When $θ = 45^{\circ}$ or $135^{\circ}$ , the value of $R$ remains the same, only the sign changes.
Reason :
$R = \frac{u^{2} \sin 2 θ}{g}$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Horizontal range, $R = \frac{u^{2} \sin 2 θ}{g}$
When $θ = 45^{\circ}$ , maximum horizontal range,
$R_{max} = \frac{u^{2}}{g} \sin 90^{\circ} = \frac{u^{2}}{g}$
When $θ = 135^{\circ}$ , maximum horizontal range,
$R = \frac{u^{2}}{g} \sin 270^{\circ} = \frac{- u^{2}}{g}$
(Negative sign implies opposite direction)

AIIMS - 2017

PHXI04:MOTION IN A PLANE

362018 For an object thrown at $45^{\circ}$ to horizontal, the maximum height ( $H$ ) and horizontal range ( $R$ ) are related as

1

R = 16 H

2

R = 8 H

3

R = 4 H

4

R = 2 H

Explanation:

For $θ = 45^{\circ}$
$H_{max} = \frac{u^{2} \sin^{2} 45^{\circ}}{2 g} = \frac{u^{2}}{4 g}$ $[∵ \sin 45^{\circ} = \frac{1}{\sqrt{2}}]$
$R = \frac{u^{2} \sin 90^{\circ}}{g} = \frac{u^{2}}{g}; ∴ \frac{R}{H} = \frac{u^{2}}{g} \times \frac{4 g}{u^{2}} = 4 \Rightarrow R = 4 H$

PHXI04:MOTION IN A PLANE

362019 Statement A :
When the range of projectile is maximum, the time of flight is the largest.
Statement B :
Range is maximum when angle of projection is $45^{\circ}$

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

$T = \frac{2 u \sin θ}{g}$ , it will maximum, when $θ = 0^{\circ}$ .
$R_{max} = \frac{u^{2}}{g}, f o r θ = 45^{\circ}$
So option (2) is correct.

PHXI04:MOTION IN A PLANE

362020 The ceiling of a long hall is 25 $m$ high. Then, the maximum horizontal distance that a ball thrown with a speed of 40 $m$ / $s$ can go without hitting the ceiling of the hall, is

1

95.5 m

2

105.5 m

3

100 m

4

150.5 m

Explanation:

Given, initial velocity $(u) = 40 m / s$
Height of the hall $(H) = 25 m$
Let the angle of projection of the ball be $θ$ , when maximum height attained by it be 25 $m$ .
Maximum height attained by the ball
$H = \frac{u^{2} \sin^{2} θ}{2 g}$
$25 = \frac{{(40)}^{2} \sin^{2} θ}{2 \times 9.8}$
or $\sin^{2} θ = \frac{25 \times 2 \times 9.8}{1600}$
$= 0.3063$
or $\sin θ = 0.5534$
$= \sin {33.6}^{\circ}$
or $θ = {33.6}^{\circ}$
$∴$ Horizontal range $(R) = \frac{u^{2} \sin 2 θ}{g}$
$= \frac{{(40)}^{2} \sin 2 \times {33.6}^{\circ}}{9.8}$
$= \frac{1600 \times \sin {67.2}^{\circ}}{9.8}$
$= \frac{1600 \times 0.9219}{9.8} = 150.5 m$

NCERT Exemplar

PHXI04:MOTION IN A PLANE

362021 The range of the projectile projected at an angle of $15^{\circ}$ with horizontal is $50 m$ . If the projectile is projected with same velocity at an angle of $45^{\circ}$ with horizontal, then its range will be

1

100 \sqrt{2} m

2

100 m

3

50 \sqrt{2} m

4

50 m

Explanation:

$R = \frac{u^{2}}{g} \sin 2 θ$
$∴ \frac{R_{45}}{R_{15}} = \frac{\sin 2 \times 45^{\circ}}{\sin 2 \times 15^{\circ}}$
$\Rightarrow R_{45} = 50 \times \frac{1}{1 / 2} m$
$= 100 m$

JEE - 2023

PHXI04:MOTION IN A PLANE

362017 Assertion :
When $θ = 45^{\circ}$ or $135^{\circ}$ , the value of $R$ remains the same, only the sign changes.
Reason :
$R = \frac{u^{2} \sin 2 θ}{g}$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Horizontal range, $R = \frac{u^{2} \sin 2 θ}{g}$
When $θ = 45^{\circ}$ , maximum horizontal range,
$R_{max} = \frac{u^{2}}{g} \sin 90^{\circ} = \frac{u^{2}}{g}$
When $θ = 135^{\circ}$ , maximum horizontal range,
$R = \frac{u^{2}}{g} \sin 270^{\circ} = \frac{- u^{2}}{g}$
(Negative sign implies opposite direction)

AIIMS - 2017

PHXI04:MOTION IN A PLANE

362018 For an object thrown at $45^{\circ}$ to horizontal, the maximum height ( $H$ ) and horizontal range ( $R$ ) are related as

1

R = 16 H

2

R = 8 H

3

R = 4 H

4

R = 2 H

Explanation:

For $θ = 45^{\circ}$
$H_{max} = \frac{u^{2} \sin^{2} 45^{\circ}}{2 g} = \frac{u^{2}}{4 g}$ $[∵ \sin 45^{\circ} = \frac{1}{\sqrt{2}}]$
$R = \frac{u^{2} \sin 90^{\circ}}{g} = \frac{u^{2}}{g}; ∴ \frac{R}{H} = \frac{u^{2}}{g} \times \frac{4 g}{u^{2}} = 4 \Rightarrow R = 4 H$

PHXI04:MOTION IN A PLANE

362019 Statement A :
When the range of projectile is maximum, the time of flight is the largest.
Statement B :
Range is maximum when angle of projection is $45^{\circ}$

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

$T = \frac{2 u \sin θ}{g}$ , it will maximum, when $θ = 0^{\circ}$ .
$R_{max} = \frac{u^{2}}{g}, f o r θ = 45^{\circ}$
So option (2) is correct.

PHXI04:MOTION IN A PLANE

362020 The ceiling of a long hall is 25 $m$ high. Then, the maximum horizontal distance that a ball thrown with a speed of 40 $m$ / $s$ can go without hitting the ceiling of the hall, is

1

95.5 m

2

105.5 m

3

100 m

4

150.5 m

Explanation:

Given, initial velocity $(u) = 40 m / s$
Height of the hall $(H) = 25 m$
Let the angle of projection of the ball be $θ$ , when maximum height attained by it be 25 $m$ .
Maximum height attained by the ball
$H = \frac{u^{2} \sin^{2} θ}{2 g}$
$25 = \frac{{(40)}^{2} \sin^{2} θ}{2 \times 9.8}$
or $\sin^{2} θ = \frac{25 \times 2 \times 9.8}{1600}$
$= 0.3063$
or $\sin θ = 0.5534$
$= \sin {33.6}^{\circ}$
or $θ = {33.6}^{\circ}$
$∴$ Horizontal range $(R) = \frac{u^{2} \sin 2 θ}{g}$
$= \frac{{(40)}^{2} \sin 2 \times {33.6}^{\circ}}{9.8}$
$= \frac{1600 \times \sin {67.2}^{\circ}}{9.8}$
$= \frac{1600 \times 0.9219}{9.8} = 150.5 m$

NCERT Exemplar

PHXI04:MOTION IN A PLANE

362021 The range of the projectile projected at an angle of $15^{\circ}$ with horizontal is $50 m$ . If the projectile is projected with same velocity at an angle of $45^{\circ}$ with horizontal, then its range will be

1

100 \sqrt{2} m

2

100 m

3

50 \sqrt{2} m

4

50 m

Explanation:

$R = \frac{u^{2}}{g} \sin 2 θ$
$∴ \frac{R_{45}}{R_{15}} = \frac{\sin 2 \times 45^{\circ}}{\sin 2 \times 15^{\circ}}$
$\Rightarrow R_{45} = 50 \times \frac{1}{1 / 2} m$
$= 100 m$

JEE - 2023

PHXI04:MOTION IN A PLANE

362017 Assertion :
When $θ = 45^{\circ}$ or $135^{\circ}$ , the value of $R$ remains the same, only the sign changes.
Reason :
$R = \frac{u^{2} \sin 2 θ}{g}$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Horizontal range, $R = \frac{u^{2} \sin 2 θ}{g}$
When $θ = 45^{\circ}$ , maximum horizontal range,
$R_{max} = \frac{u^{2}}{g} \sin 90^{\circ} = \frac{u^{2}}{g}$
When $θ = 135^{\circ}$ , maximum horizontal range,
$R = \frac{u^{2}}{g} \sin 270^{\circ} = \frac{- u^{2}}{g}$
(Negative sign implies opposite direction)

AIIMS - 2017

PHXI04:MOTION IN A PLANE

362018 For an object thrown at $45^{\circ}$ to horizontal, the maximum height ( $H$ ) and horizontal range ( $R$ ) are related as

1

R = 16 H

2

R = 8 H

3

R = 4 H

4

R = 2 H

Explanation:

For $θ = 45^{\circ}$
$H_{max} = \frac{u^{2} \sin^{2} 45^{\circ}}{2 g} = \frac{u^{2}}{4 g}$ $[∵ \sin 45^{\circ} = \frac{1}{\sqrt{2}}]$
$R = \frac{u^{2} \sin 90^{\circ}}{g} = \frac{u^{2}}{g}; ∴ \frac{R}{H} = \frac{u^{2}}{g} \times \frac{4 g}{u^{2}} = 4 \Rightarrow R = 4 H$

PHXI04:MOTION IN A PLANE

362019 Statement A :
When the range of projectile is maximum, the time of flight is the largest.
Statement B :
Range is maximum when angle of projection is $45^{\circ}$

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

$T = \frac{2 u \sin θ}{g}$ , it will maximum, when $θ = 0^{\circ}$ .
$R_{max} = \frac{u^{2}}{g}, f o r θ = 45^{\circ}$
So option (2) is correct.

PHXI04:MOTION IN A PLANE

362020 The ceiling of a long hall is 25 $m$ high. Then, the maximum horizontal distance that a ball thrown with a speed of 40 $m$ / $s$ can go without hitting the ceiling of the hall, is

1

95.5 m

2

105.5 m

3

100 m

4

150.5 m

Explanation:

Given, initial velocity $(u) = 40 m / s$
Height of the hall $(H) = 25 m$
Let the angle of projection of the ball be $θ$ , when maximum height attained by it be 25 $m$ .
Maximum height attained by the ball
$H = \frac{u^{2} \sin^{2} θ}{2 g}$
$25 = \frac{{(40)}^{2} \sin^{2} θ}{2 \times 9.8}$
or $\sin^{2} θ = \frac{25 \times 2 \times 9.8}{1600}$
$= 0.3063$
or $\sin θ = 0.5534$
$= \sin {33.6}^{\circ}$
or $θ = {33.6}^{\circ}$
$∴$ Horizontal range $(R) = \frac{u^{2} \sin 2 θ}{g}$
$= \frac{{(40)}^{2} \sin 2 \times {33.6}^{\circ}}{9.8}$
$= \frac{1600 \times \sin {67.2}^{\circ}}{9.8}$
$= \frac{1600 \times 0.9219}{9.8} = 150.5 m$

NCERT Exemplar

PHXI04:MOTION IN A PLANE

362021 The range of the projectile projected at an angle of $15^{\circ}$ with horizontal is $50 m$ . If the projectile is projected with same velocity at an angle of $45^{\circ}$ with horizontal, then its range will be

1

100 \sqrt{2} m

2

100 m

3

50 \sqrt{2} m

4

50 m

Explanation:

$R = \frac{u^{2}}{g} \sin 2 θ$
$∴ \frac{R_{45}}{R_{15}} = \frac{\sin 2 \times 45^{\circ}}{\sin 2 \times 15^{\circ}}$
$\Rightarrow R_{45} = 50 \times \frac{1}{1 / 2} m$
$= 100 m$

JEE - 2023

PHXI04:MOTION IN A PLANE

362017 Assertion :
When $θ = 45^{\circ}$ or $135^{\circ}$ , the value of $R$ remains the same, only the sign changes.
Reason :
$R = \frac{u^{2} \sin 2 θ}{g}$

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Horizontal range, $R = \frac{u^{2} \sin 2 θ}{g}$
When $θ = 45^{\circ}$ , maximum horizontal range,
$R_{max} = \frac{u^{2}}{g} \sin 90^{\circ} = \frac{u^{2}}{g}$
When $θ = 135^{\circ}$ , maximum horizontal range,
$R = \frac{u^{2}}{g} \sin 270^{\circ} = \frac{- u^{2}}{g}$
(Negative sign implies opposite direction)

AIIMS - 2017

PHXI04:MOTION IN A PLANE

362018 For an object thrown at $45^{\circ}$ to horizontal, the maximum height ( $H$ ) and horizontal range ( $R$ ) are related as

1

R = 16 H

2

R = 8 H

3

R = 4 H

4

R = 2 H

Explanation:

For $θ = 45^{\circ}$
$H_{max} = \frac{u^{2} \sin^{2} 45^{\circ}}{2 g} = \frac{u^{2}}{4 g}$ $[∵ \sin 45^{\circ} = \frac{1}{\sqrt{2}}]$
$R = \frac{u^{2} \sin 90^{\circ}}{g} = \frac{u^{2}}{g}; ∴ \frac{R}{H} = \frac{u^{2}}{g} \times \frac{4 g}{u^{2}} = 4 \Rightarrow R = 4 H$

PHXI04:MOTION IN A PLANE

362019 Statement A :
When the range of projectile is maximum, the time of flight is the largest.
Statement B :
Range is maximum when angle of projection is $45^{\circ}$

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both Statements are correct.

4 Both Statements are incorrect.

Explanation:

$T = \frac{2 u \sin θ}{g}$ , it will maximum, when $θ = 0^{\circ}$ .
$R_{max} = \frac{u^{2}}{g}, f o r θ = 45^{\circ}$
So option (2) is correct.

PHXI04:MOTION IN A PLANE

362020 The ceiling of a long hall is 25 $m$ high. Then, the maximum horizontal distance that a ball thrown with a speed of 40 $m$ / $s$ can go without hitting the ceiling of the hall, is

1

95.5 m

2

105.5 m

3

100 m

4

150.5 m

Explanation:

Given, initial velocity $(u) = 40 m / s$
Height of the hall $(H) = 25 m$
Let the angle of projection of the ball be $θ$ , when maximum height attained by it be 25 $m$ .
Maximum height attained by the ball
$H = \frac{u^{2} \sin^{2} θ}{2 g}$
$25 = \frac{{(40)}^{2} \sin^{2} θ}{2 \times 9.8}$
or $\sin^{2} θ = \frac{25 \times 2 \times 9.8}{1600}$
$= 0.3063$
or $\sin θ = 0.5534$
$= \sin {33.6}^{\circ}$
or $θ = {33.6}^{\circ}$
$∴$ Horizontal range $(R) = \frac{u^{2} \sin 2 θ}{g}$
$= \frac{{(40)}^{2} \sin 2 \times {33.6}^{\circ}}{9.8}$
$= \frac{1600 \times \sin {67.2}^{\circ}}{9.8}$
$= \frac{1600 \times 0.9219}{9.8} = 150.5 m$

NCERT Exemplar

PHXI04:MOTION IN A PLANE

362021 The range of the projectile projected at an angle of $15^{\circ}$ with horizontal is $50 m$ . If the projectile is projected with same velocity at an angle of $45^{\circ}$ with horizontal, then its range will be

1

100 \sqrt{2} m

2

100 m

3

50 \sqrt{2} m

4

50 m

Explanation:

$R = \frac{u^{2}}{g} \sin 2 θ$
$∴ \frac{R_{45}}{R_{15}} = \frac{\sin 2 \times 45^{\circ}}{\sin 2 \times 15^{\circ}}$
$\Rightarrow R_{45} = 50 \times \frac{1}{1 / 2} m$
$= 100 m$

JEE - 2023

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