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PHXI04:MOTION IN A PLANE

361937 Two bodies of different masses are released from the top of a tower. One is thrown in horizontal direction while other is dropped. Then which is reaching the ground first?

1 The body which has been thrown horizontally

2 The body which has been dropped

3 Both are reaching the ground simultaneously

4 Depends on the velocity with which the 1st body has been projected horizontally

PHXI04:MOTION IN A PLANE

361938 Water is flowing from a horizontal pipe fixed at a height of 2 $m$ from the ground. If it falls at a horizontal distance of 3 $m$ as shown in figure, the speed of water when it leaves the pipe is $(t a k e g = 9.8 m / s^{- 2})$
supporting img

1

2.4 m s^{- 1}

2

4.7 m s^{- 1}

3

7.4 m s^{- 1}

4

6.2 m s^{- 1}

Explanation:

$t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{4}{(9.8)}} = 0.64 s$
Now, $v = \frac{3}{t} = \frac{3}{0.64} = 4.7 m / s$

PHXI04:MOTION IN A PLANE

361939 A staircase contains three steps each $10 c m$ high and $20 c m$ wide as shown in figure. The minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane is
supporting img

1

1 m / s

2

2 m / s

3

3 m / s

4

4 m / s

Explanation:

For the minimum value of the horizontal velocity, the ball should just pass the edge of the $(n - 1)^{th}$ step, if it is to strike the $n^{th}$ step.
supporting img
$O$ to $B : x = 2 b = 2 \times 0.2 = 0.4$
$y = 2 h = 2 \times 0.1 = 0.2 m$
$y = \frac{g x^{2}}{2 u^{2}} \Rightarrow u = \sqrt{\frac{g x^{2}}{2 y}} = x \sqrt{\frac{g}{2 y}}$
$= (0.4) \sqrt{\frac{10}{2 \times 0.2}} = 0.4 \times 5 = 2 m / s = u_{min}$

PHXI04:MOTION IN A PLANE

361940 An aeroplane is flying horizontally with a velocity of $600 {kmh}^{- 1}$ and at a height of $1960 m$ When it is vertically above a point $A$ on the ground, a bomb is released from it. The bomb strikes the ground at point $B$ . The distance $A B$ is

1

1200 k m

2

0.33 k m

3

33.3 k m

4

3.33 k m

Explanation:

From equation of motion,
$h = \frac{1}{2} g t^{2}$
Given, $u = 600 k m h^{- 1}$ and $h = 1960 m$
$\Rightarrow t = \sqrt{\frac{2 \times 1960}{9.8}} = 20 s$
$∴ s = u t$
$= 600 \times \frac{20}{60 \times 60} = 3.33 k m$

PHXI04:MOTION IN A PLANE

361941 A body is projected downdwards at an angle of $30^{\circ}$ to the horizontal with a velocity of 9.8 $m$ / $s$ from the top of a tower 29.4 $m$ high. How long will it take before striking the ground?

1

2 s

2

1 s

3

4 s

4

3 s

Explanation:

The situation is shown in the adjoining figure.
$u \sin θ = 9.8 \sin 30^{\circ} = 4.9 m / s$
From the relation, $h = u \sin θ t + (1 / 2) g t^{2}$
We get, $29.4 = 4.9 t + (1 / 2) \times 9.8 t^{2} \Rightarrow t = 2 \sec$
supporting img

PHXI04:MOTION IN A PLANE

361937 Two bodies of different masses are released from the top of a tower. One is thrown in horizontal direction while other is dropped. Then which is reaching the ground first?

1 The body which has been thrown horizontally

2 The body which has been dropped

3 Both are reaching the ground simultaneously

4 Depends on the velocity with which the 1st body has been projected horizontally

PHXI04:MOTION IN A PLANE

361938 Water is flowing from a horizontal pipe fixed at a height of 2 $m$ from the ground. If it falls at a horizontal distance of 3 $m$ as shown in figure, the speed of water when it leaves the pipe is $(t a k e g = 9.8 m / s^{- 2})$
supporting img

1

2.4 m s^{- 1}

2

4.7 m s^{- 1}

3

7.4 m s^{- 1}

4

6.2 m s^{- 1}

Explanation:

$t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{4}{(9.8)}} = 0.64 s$
Now, $v = \frac{3}{t} = \frac{3}{0.64} = 4.7 m / s$

PHXI04:MOTION IN A PLANE

361939 A staircase contains three steps each $10 c m$ high and $20 c m$ wide as shown in figure. The minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane is
supporting img

1

1 m / s

2

2 m / s

3

3 m / s

4

4 m / s

Explanation:

For the minimum value of the horizontal velocity, the ball should just pass the edge of the $(n - 1)^{th}$ step, if it is to strike the $n^{th}$ step.
supporting img
$O$ to $B : x = 2 b = 2 \times 0.2 = 0.4$
$y = 2 h = 2 \times 0.1 = 0.2 m$
$y = \frac{g x^{2}}{2 u^{2}} \Rightarrow u = \sqrt{\frac{g x^{2}}{2 y}} = x \sqrt{\frac{g}{2 y}}$
$= (0.4) \sqrt{\frac{10}{2 \times 0.2}} = 0.4 \times 5 = 2 m / s = u_{min}$

PHXI04:MOTION IN A PLANE

361940 An aeroplane is flying horizontally with a velocity of $600 {kmh}^{- 1}$ and at a height of $1960 m$ When it is vertically above a point $A$ on the ground, a bomb is released from it. The bomb strikes the ground at point $B$ . The distance $A B$ is

1

1200 k m

2

0.33 k m

3

33.3 k m

4

3.33 k m

Explanation:

From equation of motion,
$h = \frac{1}{2} g t^{2}$
Given, $u = 600 k m h^{- 1}$ and $h = 1960 m$
$\Rightarrow t = \sqrt{\frac{2 \times 1960}{9.8}} = 20 s$
$∴ s = u t$
$= 600 \times \frac{20}{60 \times 60} = 3.33 k m$

PHXI04:MOTION IN A PLANE

361941 A body is projected downdwards at an angle of $30^{\circ}$ to the horizontal with a velocity of 9.8 $m$ / $s$ from the top of a tower 29.4 $m$ high. How long will it take before striking the ground?

1

2 s

2

1 s

3

4 s

4

3 s

Explanation:

The situation is shown in the adjoining figure.
$u \sin θ = 9.8 \sin 30^{\circ} = 4.9 m / s$
From the relation, $h = u \sin θ t + (1 / 2) g t^{2}$
We get, $29.4 = 4.9 t + (1 / 2) \times 9.8 t^{2} \Rightarrow t = 2 \sec$
supporting img

PHXI04:MOTION IN A PLANE

361937 Two bodies of different masses are released from the top of a tower. One is thrown in horizontal direction while other is dropped. Then which is reaching the ground first?

1 The body which has been thrown horizontally

2 The body which has been dropped

3 Both are reaching the ground simultaneously

4 Depends on the velocity with which the 1st body has been projected horizontally

PHXI04:MOTION IN A PLANE

361938 Water is flowing from a horizontal pipe fixed at a height of 2 $m$ from the ground. If it falls at a horizontal distance of 3 $m$ as shown in figure, the speed of water when it leaves the pipe is $(t a k e g = 9.8 m / s^{- 2})$
supporting img

1

2.4 m s^{- 1}

2

4.7 m s^{- 1}

3

7.4 m s^{- 1}

4

6.2 m s^{- 1}

Explanation:

$t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{4}{(9.8)}} = 0.64 s$
Now, $v = \frac{3}{t} = \frac{3}{0.64} = 4.7 m / s$

PHXI04:MOTION IN A PLANE

361939 A staircase contains three steps each $10 c m$ high and $20 c m$ wide as shown in figure. The minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane is
supporting img

1

1 m / s

2

2 m / s

3

3 m / s

4

4 m / s

Explanation:

For the minimum value of the horizontal velocity, the ball should just pass the edge of the $(n - 1)^{th}$ step, if it is to strike the $n^{th}$ step.
supporting img
$O$ to $B : x = 2 b = 2 \times 0.2 = 0.4$
$y = 2 h = 2 \times 0.1 = 0.2 m$
$y = \frac{g x^{2}}{2 u^{2}} \Rightarrow u = \sqrt{\frac{g x^{2}}{2 y}} = x \sqrt{\frac{g}{2 y}}$
$= (0.4) \sqrt{\frac{10}{2 \times 0.2}} = 0.4 \times 5 = 2 m / s = u_{min}$

PHXI04:MOTION IN A PLANE

361940 An aeroplane is flying horizontally with a velocity of $600 {kmh}^{- 1}$ and at a height of $1960 m$ When it is vertically above a point $A$ on the ground, a bomb is released from it. The bomb strikes the ground at point $B$ . The distance $A B$ is

1

1200 k m

2

0.33 k m

3

33.3 k m

4

3.33 k m

Explanation:

From equation of motion,
$h = \frac{1}{2} g t^{2}$
Given, $u = 600 k m h^{- 1}$ and $h = 1960 m$
$\Rightarrow t = \sqrt{\frac{2 \times 1960}{9.8}} = 20 s$
$∴ s = u t$
$= 600 \times \frac{20}{60 \times 60} = 3.33 k m$

PHXI04:MOTION IN A PLANE

361941 A body is projected downdwards at an angle of $30^{\circ}$ to the horizontal with a velocity of 9.8 $m$ / $s$ from the top of a tower 29.4 $m$ high. How long will it take before striking the ground?

1

2 s

2

1 s

3

4 s

4

3 s

Explanation:

The situation is shown in the adjoining figure.
$u \sin θ = 9.8 \sin 30^{\circ} = 4.9 m / s$
From the relation, $h = u \sin θ t + (1 / 2) g t^{2}$
We get, $29.4 = 4.9 t + (1 / 2) \times 9.8 t^{2} \Rightarrow t = 2 \sec$
supporting img

PHXI04:MOTION IN A PLANE

361937 Two bodies of different masses are released from the top of a tower. One is thrown in horizontal direction while other is dropped. Then which is reaching the ground first?

1 The body which has been thrown horizontally

2 The body which has been dropped

3 Both are reaching the ground simultaneously

4 Depends on the velocity with which the 1st body has been projected horizontally

PHXI04:MOTION IN A PLANE

361938 Water is flowing from a horizontal pipe fixed at a height of 2 $m$ from the ground. If it falls at a horizontal distance of 3 $m$ as shown in figure, the speed of water when it leaves the pipe is $(t a k e g = 9.8 m / s^{- 2})$
supporting img

1

2.4 m s^{- 1}

2

4.7 m s^{- 1}

3

7.4 m s^{- 1}

4

6.2 m s^{- 1}

Explanation:

$t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{4}{(9.8)}} = 0.64 s$
Now, $v = \frac{3}{t} = \frac{3}{0.64} = 4.7 m / s$

PHXI04:MOTION IN A PLANE

361939 A staircase contains three steps each $10 c m$ high and $20 c m$ wide as shown in figure. The minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane is
supporting img

1

1 m / s

2

2 m / s

3

3 m / s

4

4 m / s

Explanation:

For the minimum value of the horizontal velocity, the ball should just pass the edge of the $(n - 1)^{th}$ step, if it is to strike the $n^{th}$ step.
supporting img
$O$ to $B : x = 2 b = 2 \times 0.2 = 0.4$
$y = 2 h = 2 \times 0.1 = 0.2 m$
$y = \frac{g x^{2}}{2 u^{2}} \Rightarrow u = \sqrt{\frac{g x^{2}}{2 y}} = x \sqrt{\frac{g}{2 y}}$
$= (0.4) \sqrt{\frac{10}{2 \times 0.2}} = 0.4 \times 5 = 2 m / s = u_{min}$

PHXI04:MOTION IN A PLANE

361940 An aeroplane is flying horizontally with a velocity of $600 {kmh}^{- 1}$ and at a height of $1960 m$ When it is vertically above a point $A$ on the ground, a bomb is released from it. The bomb strikes the ground at point $B$ . The distance $A B$ is

1

1200 k m

2

0.33 k m

3

33.3 k m

4

3.33 k m

Explanation:

From equation of motion,
$h = \frac{1}{2} g t^{2}$
Given, $u = 600 k m h^{- 1}$ and $h = 1960 m$
$\Rightarrow t = \sqrt{\frac{2 \times 1960}{9.8}} = 20 s$
$∴ s = u t$
$= 600 \times \frac{20}{60 \times 60} = 3.33 k m$

PHXI04:MOTION IN A PLANE

361941 A body is projected downdwards at an angle of $30^{\circ}$ to the horizontal with a velocity of 9.8 $m$ / $s$ from the top of a tower 29.4 $m$ high. How long will it take before striking the ground?

1

2 s

2

1 s

3

4 s

4

3 s

Explanation:

The situation is shown in the adjoining figure.
$u \sin θ = 9.8 \sin 30^{\circ} = 4.9 m / s$
From the relation, $h = u \sin θ t + (1 / 2) g t^{2}$
We get, $29.4 = 4.9 t + (1 / 2) \times 9.8 t^{2} \Rightarrow t = 2 \sec$
supporting img

PHXI04:MOTION IN A PLANE

361937 Two bodies of different masses are released from the top of a tower. One is thrown in horizontal direction while other is dropped. Then which is reaching the ground first?

1 The body which has been thrown horizontally

2 The body which has been dropped

3 Both are reaching the ground simultaneously

4 Depends on the velocity with which the 1st body has been projected horizontally

PHXI04:MOTION IN A PLANE

361938 Water is flowing from a horizontal pipe fixed at a height of 2 $m$ from the ground. If it falls at a horizontal distance of 3 $m$ as shown in figure, the speed of water when it leaves the pipe is $(t a k e g = 9.8 m / s^{- 2})$
supporting img

1

2.4 m s^{- 1}

2

4.7 m s^{- 1}

3

7.4 m s^{- 1}

4

6.2 m s^{- 1}

Explanation:

$t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{4}{(9.8)}} = 0.64 s$
Now, $v = \frac{3}{t} = \frac{3}{0.64} = 4.7 m / s$

PHXI04:MOTION IN A PLANE

361939 A staircase contains three steps each $10 c m$ high and $20 c m$ wide as shown in figure. The minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane is
supporting img

1

1 m / s

2

2 m / s

3

3 m / s

4

4 m / s

Explanation:

For the minimum value of the horizontal velocity, the ball should just pass the edge of the $(n - 1)^{th}$ step, if it is to strike the $n^{th}$ step.
supporting img
$O$ to $B : x = 2 b = 2 \times 0.2 = 0.4$
$y = 2 h = 2 \times 0.1 = 0.2 m$
$y = \frac{g x^{2}}{2 u^{2}} \Rightarrow u = \sqrt{\frac{g x^{2}}{2 y}} = x \sqrt{\frac{g}{2 y}}$
$= (0.4) \sqrt{\frac{10}{2 \times 0.2}} = 0.4 \times 5 = 2 m / s = u_{min}$

PHXI04:MOTION IN A PLANE

361940 An aeroplane is flying horizontally with a velocity of $600 {kmh}^{- 1}$ and at a height of $1960 m$ When it is vertically above a point $A$ on the ground, a bomb is released from it. The bomb strikes the ground at point $B$ . The distance $A B$ is

1

1200 k m

2

0.33 k m

3

33.3 k m

4

3.33 k m

Explanation:

From equation of motion,
$h = \frac{1}{2} g t^{2}$
Given, $u = 600 k m h^{- 1}$ and $h = 1960 m$
$\Rightarrow t = \sqrt{\frac{2 \times 1960}{9.8}} = 20 s$
$∴ s = u t$
$= 600 \times \frac{20}{60 \times 60} = 3.33 k m$

PHXI04:MOTION IN A PLANE

361941 A body is projected downdwards at an angle of $30^{\circ}$ to the horizontal with a velocity of 9.8 $m$ / $s$ from the top of a tower 29.4 $m$ high. How long will it take before striking the ground?

1

2 s

2

1 s

3

4 s

4

3 s

Explanation:

The situation is shown in the adjoining figure.
$u \sin θ = 9.8 \sin 30^{\circ} = 4.9 m / s$
From the relation, $h = u \sin θ t + (1 / 2) g t^{2}$
We get, $29.4 = 4.9 t + (1 / 2) \times 9.8 t^{2} \Rightarrow t = 2 \sec$
supporting img

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