QBManager

Motion in One Dimensions

141856 A person throws balls into air vertically upward in regular intervals of time of one second. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the balls rise is :
(Assume, $g = 10 {ms}^{- 2}$ )

1

20 m

2

5 m

3

10 m

4

7.5 m

Explanation:

B Given,
$v = 0, g = 10 m / s^{- 2}, t = 1 \sec$ .
From first equation of motion, we get-
$v = u - gt$
$0 = u - 10 \times 1$
$u = 10 m / s$
By using third equation of motion, we get
$v^{2} = u^{2} - 2 gH$
$0 = u^{2} - 2 gH$
Height $(H)$ , to which the ball will rise is,
$H = \frac{u^{2}}{2 g}$
$H = \frac{(10)^{2}}{2 \times 10}$
$H = 5 m$

Karnataka CET-2012

Motion in One Dimensions

141860 A body is released from a great height falls freely towards the earth. Another body is released from the same height exactly a second later. Then the separation between two bodies, $2$ sec after the release of the second body is nearly

1

15 m

2

20 m

3

25 m

4

30 m

Explanation:

C Given that,
Initial velocity $u = 0$
Time for the first body $(t_{1}) = 3$ sec
Time for the second body $(t_{2}) = 2 \sec$
The first body fall through a height $s_{1}$ during $3 \sec$ .
$s_{1} = 0 + \frac{1}{2} g (3)^{2}$
$s_{1} = 4.5 \times 9.8 m = 44.1 m$
The second body fall through a height $s_{2}$ during $2 \sec$ .
$s_{2} = 0 + \frac{1}{2} \times 9.8 \times 2^{2}$
$s_{2} = 2 \times 9.8 = 19.6 m$
Separation between both bodies, we get-
$s = s_{1} - s_{2}$
$s = 44.1 - 19.6 \Rightarrow s = 24.5 m ≃ 25 m$

J and K-CET- 2007

Motion in One Dimensions

141862 A ball released at rest from the top of a tower reaches the ground in $4 s$ . The height of the ball from the ground at $t = 2 s$ of its journey is

1

60 m

2

40 m

3

20 m

4

15 m

Explanation:

A Given that,
$t_{1} = 4 \sec$
$u = 0$
$t_{2} = 2 \sec$
Let $h$ be the height of the tower
As the ball reaches the ground in $4 s$ ,
$∴ h_{1} = 0 + \frac{1}{2} {gt}^{2}$
$h_{1} = \frac{1}{2} \times 10 \times 4^{2}$
$h_{1} = 80 m$
The height fallen by the ball in $2 s$
$h_{2} = \frac{1}{2} \times 10 \times 2^{2}$
$h_{2} = 20 m$
The height of ball from the ground
$h^{'} = h_{1} - h_{2}$
$h^{'} = 80 - 20$
$h^{'} = 60 m$

J and K-CET-2016

Motion in One Dimensions

141863 A ball is dropped from the top of $80 m$ high tower. If after $2$ sec of fall the gravity $(g = 10$ $m / s^{2}$ ) disappears, then time taken to reach ground since the gravity disappeared is

1

2 \sec

2

3 \sec

3

4 \sec

4

5 \sec

Explanation:

B Given,
$Height of tower (h) = 80 m$
Time $(t) = 2 \sec$ .
Initial velocity $(u) = 0$
original image
So,
First equation of motion -
$v = u + gt$
$v = 0 + 10 \times 2$
$v = 20 m / s$
Distance travelled in $2 \sec$
$h = ut + \frac{1}{2} {at}^{2}$
$h = \frac{1}{2} \times 10 \times 2^{2}$
$h = 20 m$
Distance to reach ground,
$d = H - h$
$d = 80 - 20$
$d = 60 m$
Time,
$t = \frac{d}{v}$
$t = \frac{60}{20} = 3 \sec$

J and K-CET-2015

Motion in One Dimensions

141856 A person throws balls into air vertically upward in regular intervals of time of one second. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the balls rise is :
(Assume, $g = 10 {ms}^{- 2}$ )

1

20 m

2

5 m

3

10 m

4

7.5 m

Explanation:

B Given,
$v = 0, g = 10 m / s^{- 2}, t = 1 \sec$ .
From first equation of motion, we get-
$v = u - gt$
$0 = u - 10 \times 1$
$u = 10 m / s$
By using third equation of motion, we get
$v^{2} = u^{2} - 2 gH$
$0 = u^{2} - 2 gH$
Height $(H)$ , to which the ball will rise is,
$H = \frac{u^{2}}{2 g}$
$H = \frac{(10)^{2}}{2 \times 10}$
$H = 5 m$

Karnataka CET-2012

Motion in One Dimensions

141860 A body is released from a great height falls freely towards the earth. Another body is released from the same height exactly a second later. Then the separation between two bodies, $2$ sec after the release of the second body is nearly

1

15 m

2

20 m

3

25 m

4

30 m

Explanation:

C Given that,
Initial velocity $u = 0$
Time for the first body $(t_{1}) = 3$ sec
Time for the second body $(t_{2}) = 2 \sec$
The first body fall through a height $s_{1}$ during $3 \sec$ .
$s_{1} = 0 + \frac{1}{2} g (3)^{2}$
$s_{1} = 4.5 \times 9.8 m = 44.1 m$
The second body fall through a height $s_{2}$ during $2 \sec$ .
$s_{2} = 0 + \frac{1}{2} \times 9.8 \times 2^{2}$
$s_{2} = 2 \times 9.8 = 19.6 m$
Separation between both bodies, we get-
$s = s_{1} - s_{2}$
$s = 44.1 - 19.6 \Rightarrow s = 24.5 m ≃ 25 m$

J and K-CET- 2007

Motion in One Dimensions

141862 A ball released at rest from the top of a tower reaches the ground in $4 s$ . The height of the ball from the ground at $t = 2 s$ of its journey is

1

60 m

2

40 m

3

20 m

4

15 m

Explanation:

A Given that,
$t_{1} = 4 \sec$
$u = 0$
$t_{2} = 2 \sec$
Let $h$ be the height of the tower
As the ball reaches the ground in $4 s$ ,
$∴ h_{1} = 0 + \frac{1}{2} {gt}^{2}$
$h_{1} = \frac{1}{2} \times 10 \times 4^{2}$
$h_{1} = 80 m$
The height fallen by the ball in $2 s$
$h_{2} = \frac{1}{2} \times 10 \times 2^{2}$
$h_{2} = 20 m$
The height of ball from the ground
$h^{'} = h_{1} - h_{2}$
$h^{'} = 80 - 20$
$h^{'} = 60 m$

J and K-CET-2016

Motion in One Dimensions

141863 A ball is dropped from the top of $80 m$ high tower. If after $2$ sec of fall the gravity $(g = 10$ $m / s^{2}$ ) disappears, then time taken to reach ground since the gravity disappeared is

1

2 \sec

2

3 \sec

3

4 \sec

4

5 \sec

Explanation:

B Given,
$Height of tower (h) = 80 m$
Time $(t) = 2 \sec$ .
Initial velocity $(u) = 0$
original image
So,
First equation of motion -
$v = u + gt$
$v = 0 + 10 \times 2$
$v = 20 m / s$
Distance travelled in $2 \sec$
$h = ut + \frac{1}{2} {at}^{2}$
$h = \frac{1}{2} \times 10 \times 2^{2}$
$h = 20 m$
Distance to reach ground,
$d = H - h$
$d = 80 - 20$
$d = 60 m$
Time,
$t = \frac{d}{v}$
$t = \frac{60}{20} = 3 \sec$

J and K-CET-2015

Motion in One Dimensions

141856 A person throws balls into air vertically upward in regular intervals of time of one second. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the balls rise is :
(Assume, $g = 10 {ms}^{- 2}$ )

1

20 m

2

5 m

3

10 m

4

7.5 m

Explanation:

B Given,
$v = 0, g = 10 m / s^{- 2}, t = 1 \sec$ .
From first equation of motion, we get-
$v = u - gt$
$0 = u - 10 \times 1$
$u = 10 m / s$
By using third equation of motion, we get
$v^{2} = u^{2} - 2 gH$
$0 = u^{2} - 2 gH$
Height $(H)$ , to which the ball will rise is,
$H = \frac{u^{2}}{2 g}$
$H = \frac{(10)^{2}}{2 \times 10}$
$H = 5 m$

Karnataka CET-2012

Motion in One Dimensions

141860 A body is released from a great height falls freely towards the earth. Another body is released from the same height exactly a second later. Then the separation between two bodies, $2$ sec after the release of the second body is nearly

1

15 m

2

20 m

3

25 m

4

30 m

Explanation:

C Given that,
Initial velocity $u = 0$
Time for the first body $(t_{1}) = 3$ sec
Time for the second body $(t_{2}) = 2 \sec$
The first body fall through a height $s_{1}$ during $3 \sec$ .
$s_{1} = 0 + \frac{1}{2} g (3)^{2}$
$s_{1} = 4.5 \times 9.8 m = 44.1 m$
The second body fall through a height $s_{2}$ during $2 \sec$ .
$s_{2} = 0 + \frac{1}{2} \times 9.8 \times 2^{2}$
$s_{2} = 2 \times 9.8 = 19.6 m$
Separation between both bodies, we get-
$s = s_{1} - s_{2}$
$s = 44.1 - 19.6 \Rightarrow s = 24.5 m ≃ 25 m$

J and K-CET- 2007

Motion in One Dimensions

141862 A ball released at rest from the top of a tower reaches the ground in $4 s$ . The height of the ball from the ground at $t = 2 s$ of its journey is

1

60 m

2

40 m

3

20 m

4

15 m

Explanation:

A Given that,
$t_{1} = 4 \sec$
$u = 0$
$t_{2} = 2 \sec$
Let $h$ be the height of the tower
As the ball reaches the ground in $4 s$ ,
$∴ h_{1} = 0 + \frac{1}{2} {gt}^{2}$
$h_{1} = \frac{1}{2} \times 10 \times 4^{2}$
$h_{1} = 80 m$
The height fallen by the ball in $2 s$
$h_{2} = \frac{1}{2} \times 10 \times 2^{2}$
$h_{2} = 20 m$
The height of ball from the ground
$h^{'} = h_{1} - h_{2}$
$h^{'} = 80 - 20$
$h^{'} = 60 m$

J and K-CET-2016

Motion in One Dimensions

141863 A ball is dropped from the top of $80 m$ high tower. If after $2$ sec of fall the gravity $(g = 10$ $m / s^{2}$ ) disappears, then time taken to reach ground since the gravity disappeared is

1

2 \sec

2

3 \sec

3

4 \sec

4

5 \sec

Explanation:

B Given,
$Height of tower (h) = 80 m$
Time $(t) = 2 \sec$ .
Initial velocity $(u) = 0$
original image
So,
First equation of motion -
$v = u + gt$
$v = 0 + 10 \times 2$
$v = 20 m / s$
Distance travelled in $2 \sec$
$h = ut + \frac{1}{2} {at}^{2}$
$h = \frac{1}{2} \times 10 \times 2^{2}$
$h = 20 m$
Distance to reach ground,
$d = H - h$
$d = 80 - 20$
$d = 60 m$
Time,
$t = \frac{d}{v}$
$t = \frac{60}{20} = 3 \sec$

J and K-CET-2015

Motion in One Dimensions

141856 A person throws balls into air vertically upward in regular intervals of time of one second. The next ball is thrown when the velocity of the ball thrown earlier becomes zero. The height to which the balls rise is :
(Assume, $g = 10 {ms}^{- 2}$ )

1

20 m

2

5 m

3

10 m

4

7.5 m

Explanation:

B Given,
$v = 0, g = 10 m / s^{- 2}, t = 1 \sec$ .
From first equation of motion, we get-
$v = u - gt$
$0 = u - 10 \times 1$
$u = 10 m / s$
By using third equation of motion, we get
$v^{2} = u^{2} - 2 gH$
$0 = u^{2} - 2 gH$
Height $(H)$ , to which the ball will rise is,
$H = \frac{u^{2}}{2 g}$
$H = \frac{(10)^{2}}{2 \times 10}$
$H = 5 m$

Karnataka CET-2012

Motion in One Dimensions

141860 A body is released from a great height falls freely towards the earth. Another body is released from the same height exactly a second later. Then the separation between two bodies, $2$ sec after the release of the second body is nearly

1

15 m

2

20 m

3

25 m

4

30 m

Explanation:

C Given that,
Initial velocity $u = 0$
Time for the first body $(t_{1}) = 3$ sec
Time for the second body $(t_{2}) = 2 \sec$
The first body fall through a height $s_{1}$ during $3 \sec$ .
$s_{1} = 0 + \frac{1}{2} g (3)^{2}$
$s_{1} = 4.5 \times 9.8 m = 44.1 m$
The second body fall through a height $s_{2}$ during $2 \sec$ .
$s_{2} = 0 + \frac{1}{2} \times 9.8 \times 2^{2}$
$s_{2} = 2 \times 9.8 = 19.6 m$
Separation between both bodies, we get-
$s = s_{1} - s_{2}$
$s = 44.1 - 19.6 \Rightarrow s = 24.5 m ≃ 25 m$

J and K-CET- 2007

Motion in One Dimensions

141862 A ball released at rest from the top of a tower reaches the ground in $4 s$ . The height of the ball from the ground at $t = 2 s$ of its journey is

1

60 m

2

40 m

3

20 m

4

15 m

Explanation:

A Given that,
$t_{1} = 4 \sec$
$u = 0$
$t_{2} = 2 \sec$
Let $h$ be the height of the tower
As the ball reaches the ground in $4 s$ ,
$∴ h_{1} = 0 + \frac{1}{2} {gt}^{2}$
$h_{1} = \frac{1}{2} \times 10 \times 4^{2}$
$h_{1} = 80 m$
The height fallen by the ball in $2 s$
$h_{2} = \frac{1}{2} \times 10 \times 2^{2}$
$h_{2} = 20 m$
The height of ball from the ground
$h^{'} = h_{1} - h_{2}$
$h^{'} = 80 - 20$
$h^{'} = 60 m$

J and K-CET-2016

Motion in One Dimensions

141863 A ball is dropped from the top of $80 m$ high tower. If after $2$ sec of fall the gravity $(g = 10$ $m / s^{2}$ ) disappears, then time taken to reach ground since the gravity disappeared is

1

2 \sec

2

3 \sec

3

4 \sec

4

5 \sec

Explanation:

B Given,
$Height of tower (h) = 80 m$
Time $(t) = 2 \sec$ .
Initial velocity $(u) = 0$
original image
So,
First equation of motion -
$v = u + gt$
$v = 0 + 10 \times 2$
$v = 20 m / s$
Distance travelled in $2 \sec$
$h = ut + \frac{1}{2} {at}^{2}$
$h = \frac{1}{2} \times 10 \times 2^{2}$
$h = 20 m$
Distance to reach ground,
$d = H - h$
$d = 80 - 20$
$d = 60 m$
Time,
$t = \frac{d}{v}$
$t = \frac{60}{20} = 3 \sec$

J and K-CET-2015

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