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PHXI04:MOTION IN A PLANE

362030 Assertion :
The height attained by a projectile is twenty five percent of range, when projected for maximum range.
Reason :
The maximum height of a projectile is independent of initial velocity of projectile.

1 Both assertion and reason are correct and reason isthe correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Range will be maximum when $θ = 45^{\circ}$ and in this condition $R = 4 H \Rightarrow H = \frac{R}{4}$ (always)
So maximum height is 25% of maximum range.Maximum height of a projectile is $H = \frac{u^{2} \sin^{2} θ}{2 g}$ . $H$ depends on initial velocity. Option (3) is correct.

PHXI04:MOTION IN A PLANE

362031 When projected at an angle of $15^{\circ}$ , the range of a projectile is $50 m$ . If with the same speed, it is projected at an angle of $45^{\circ}$ , its range will be

1

150 m

2

100 m

3

50 m

4

25 m

Explanation:

Given : $θ_{1} = 15^{\circ}, R_{1} = 50 m, θ_{2} = 45^{\circ}, R =$ ? Range of a projectile is $R = \frac{u^{2} \sin 2 θ}{g}$
$50 = \frac{u^{2} \sin 30^{\circ}}{g} (1)$
$R_{2} = \frac{u^{2} \sin 90^{\circ}}{g} (2)$
Dividing eq. (1) by eq. (2), we get $\frac{50}{R_{2}} = \frac{1}{2} \Rightarrow R_{2} = 100 m$ .
So correct option is (2)

PHXI04:MOTION IN A PLANE

362032 A grasshopper can jump maximum distance 0.8 $m$ , while spending negligible time on the ground. How far can it go in $3 s (g = 10 m s^{- 2})$ ?

1

6 m

2

4 m

3

8 m

4

10 m

Explanation:

Maximum range $(θ = 45^{\circ}) = \frac{u^{2}}{g} = 0.8 u = 2 \sqrt{2} m$
Distance covered in $3 s = (4 \cos 45^{\circ}) (3)$
$= (2 \sqrt{2}) (\frac{1}{\sqrt{2}}) (3) = 6 m$

PHXI04:MOTION IN A PLANE

362033 Two objects are projected with same velocity ' $u$ ' however at different angles $α$ and $β$ with the horizontal. If $α + β = 90^{\circ}$ , the ratio of horizontal range of the first object to the $2^{n d}$ object will be

1

1 : 2

2

2 : 1

3

4 : 1

4

1 : 1

Explanation:

Horizontal range is same for two complementary angles, so, $R_{1} : R_{2} = 1 : 1$ .

JEE - 2023

PHXI04:MOTION IN A PLANE

362030 Assertion :
The height attained by a projectile is twenty five percent of range, when projected for maximum range.
Reason :
The maximum height of a projectile is independent of initial velocity of projectile.

1 Both assertion and reason are correct and reason isthe correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Range will be maximum when $θ = 45^{\circ}$ and in this condition $R = 4 H \Rightarrow H = \frac{R}{4}$ (always)
So maximum height is 25% of maximum range.Maximum height of a projectile is $H = \frac{u^{2} \sin^{2} θ}{2 g}$ . $H$ depends on initial velocity. Option (3) is correct.

PHXI04:MOTION IN A PLANE

362031 When projected at an angle of $15^{\circ}$ , the range of a projectile is $50 m$ . If with the same speed, it is projected at an angle of $45^{\circ}$ , its range will be

1

150 m

2

100 m

3

50 m

4

25 m

Explanation:

Given : $θ_{1} = 15^{\circ}, R_{1} = 50 m, θ_{2} = 45^{\circ}, R =$ ? Range of a projectile is $R = \frac{u^{2} \sin 2 θ}{g}$
$50 = \frac{u^{2} \sin 30^{\circ}}{g} (1)$
$R_{2} = \frac{u^{2} \sin 90^{\circ}}{g} (2)$
Dividing eq. (1) by eq. (2), we get $\frac{50}{R_{2}} = \frac{1}{2} \Rightarrow R_{2} = 100 m$ .
So correct option is (2)

PHXI04:MOTION IN A PLANE

362032 A grasshopper can jump maximum distance 0.8 $m$ , while spending negligible time on the ground. How far can it go in $3 s (g = 10 m s^{- 2})$ ?

1

6 m

2

4 m

3

8 m

4

10 m

Explanation:

Maximum range $(θ = 45^{\circ}) = \frac{u^{2}}{g} = 0.8 u = 2 \sqrt{2} m$
Distance covered in $3 s = (4 \cos 45^{\circ}) (3)$
$= (2 \sqrt{2}) (\frac{1}{\sqrt{2}}) (3) = 6 m$

PHXI04:MOTION IN A PLANE

362033 Two objects are projected with same velocity ' $u$ ' however at different angles $α$ and $β$ with the horizontal. If $α + β = 90^{\circ}$ , the ratio of horizontal range of the first object to the $2^{n d}$ object will be

1

1 : 2

2

2 : 1

3

4 : 1

4

1 : 1

Explanation:

Horizontal range is same for two complementary angles, so, $R_{1} : R_{2} = 1 : 1$ .

JEE - 2023

PHXI04:MOTION IN A PLANE

362030 Assertion :
The height attained by a projectile is twenty five percent of range, when projected for maximum range.
Reason :
The maximum height of a projectile is independent of initial velocity of projectile.

1 Both assertion and reason are correct and reason isthe correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Range will be maximum when $θ = 45^{\circ}$ and in this condition $R = 4 H \Rightarrow H = \frac{R}{4}$ (always)
So maximum height is 25% of maximum range.Maximum height of a projectile is $H = \frac{u^{2} \sin^{2} θ}{2 g}$ . $H$ depends on initial velocity. Option (3) is correct.

PHXI04:MOTION IN A PLANE

362031 When projected at an angle of $15^{\circ}$ , the range of a projectile is $50 m$ . If with the same speed, it is projected at an angle of $45^{\circ}$ , its range will be

1

150 m

2

100 m

3

50 m

4

25 m

Explanation:

Given : $θ_{1} = 15^{\circ}, R_{1} = 50 m, θ_{2} = 45^{\circ}, R =$ ? Range of a projectile is $R = \frac{u^{2} \sin 2 θ}{g}$
$50 = \frac{u^{2} \sin 30^{\circ}}{g} (1)$
$R_{2} = \frac{u^{2} \sin 90^{\circ}}{g} (2)$
Dividing eq. (1) by eq. (2), we get $\frac{50}{R_{2}} = \frac{1}{2} \Rightarrow R_{2} = 100 m$ .
So correct option is (2)

PHXI04:MOTION IN A PLANE

362032 A grasshopper can jump maximum distance 0.8 $m$ , while spending negligible time on the ground. How far can it go in $3 s (g = 10 m s^{- 2})$ ?

1

6 m

2

4 m

3

8 m

4

10 m

Explanation:

Maximum range $(θ = 45^{\circ}) = \frac{u^{2}}{g} = 0.8 u = 2 \sqrt{2} m$
Distance covered in $3 s = (4 \cos 45^{\circ}) (3)$
$= (2 \sqrt{2}) (\frac{1}{\sqrt{2}}) (3) = 6 m$

PHXI04:MOTION IN A PLANE

362033 Two objects are projected with same velocity ' $u$ ' however at different angles $α$ and $β$ with the horizontal. If $α + β = 90^{\circ}$ , the ratio of horizontal range of the first object to the $2^{n d}$ object will be

1

1 : 2

2

2 : 1

3

4 : 1

4

1 : 1

Explanation:

Horizontal range is same for two complementary angles, so, $R_{1} : R_{2} = 1 : 1$ .

JEE - 2023

PHXI04:MOTION IN A PLANE

362030 Assertion :
The height attained by a projectile is twenty five percent of range, when projected for maximum range.
Reason :
The maximum height of a projectile is independent of initial velocity of projectile.

1 Both assertion and reason are correct and reason isthe correct explanation of assertion.

2 Both assertion and reason are correct but reason is not the correct explanation of assertion.

3 Assertion is correct but reason is incorrect.

4 Assertion is incorrect but reason is correct.

Explanation:

Range will be maximum when $θ = 45^{\circ}$ and in this condition $R = 4 H \Rightarrow H = \frac{R}{4}$ (always)
So maximum height is 25% of maximum range.Maximum height of a projectile is $H = \frac{u^{2} \sin^{2} θ}{2 g}$ . $H$ depends on initial velocity. Option (3) is correct.

PHXI04:MOTION IN A PLANE

362031 When projected at an angle of $15^{\circ}$ , the range of a projectile is $50 m$ . If with the same speed, it is projected at an angle of $45^{\circ}$ , its range will be

1

150 m

2

100 m

3

50 m

4

25 m

Explanation:

Given : $θ_{1} = 15^{\circ}, R_{1} = 50 m, θ_{2} = 45^{\circ}, R =$ ? Range of a projectile is $R = \frac{u^{2} \sin 2 θ}{g}$
$50 = \frac{u^{2} \sin 30^{\circ}}{g} (1)$
$R_{2} = \frac{u^{2} \sin 90^{\circ}}{g} (2)$
Dividing eq. (1) by eq. (2), we get $\frac{50}{R_{2}} = \frac{1}{2} \Rightarrow R_{2} = 100 m$ .
So correct option is (2)

PHXI04:MOTION IN A PLANE

362032 A grasshopper can jump maximum distance 0.8 $m$ , while spending negligible time on the ground. How far can it go in $3 s (g = 10 m s^{- 2})$ ?

1

6 m

2

4 m

3

8 m

4

10 m

Explanation:

Maximum range $(θ = 45^{\circ}) = \frac{u^{2}}{g} = 0.8 u = 2 \sqrt{2} m$
Distance covered in $3 s = (4 \cos 45^{\circ}) (3)$
$= (2 \sqrt{2}) (\frac{1}{\sqrt{2}}) (3) = 6 m$

PHXI04:MOTION IN A PLANE

362033 Two objects are projected with same velocity ' $u$ ' however at different angles $α$ and $β$ with the horizontal. If $α + β = 90^{\circ}$ , the ratio of horizontal range of the first object to the $2^{n d}$ object will be

1

1 : 2

2

2 : 1

3

4 : 1

4

1 : 1

Explanation:

Horizontal range is same for two complementary angles, so, $R_{1} : R_{2} = 1 : 1$ .

JEE - 2023

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