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PHXI03:MOTION IN A STRAIGHT LINE

362349 A ball is dropped from a height of 19.6 $m$ . The distance covered by it in the last second is

1 19.6

m

2 14.7

m

3 4.8

m

4 9.8

m

Explanation:

Height, $h = 19.6 m$
Time taken $(t)$ to cover this distance will be given by
$s = u t + \frac{1}{2} a t^{2}$
Here, initial velocity, $u = 0$
$∴ h = \frac{1}{2} g t^{2} \Rightarrow t = \sqrt{\frac{2 h}{g}}$
$\Rightarrow t = \sqrt{\frac{2 \times 19.6}{9.8}} = 2 s$
Velocity acquired in first one second of the motion
$v = u + a t = v = 9.8 \times 1 m / s = 9.8 m / s$
This will be the initial velocity $(u^{'})$ for the last one second of the motion.
$∴$ the distance covered in last second will be
$s = u^{'} t + \frac{1}{2} g + t^{2}$
$\Rightarrow s = 9.8 \times 1 + \frac{1}{2} \times 9.8 \times 1^{2}$
$s = 9.8 + 4.9 = 14.7 m$ .
So correct option is (2)

PHXI03:MOTION IN A STRAIGHT LINE

362350 A parachutist after bailing out falls $50 m$ without friction. When parachute opens, it decelerates at $2 m s^{- 2}$ . He reaches the ground with a speed of $3 m s^{- 1}$ . At what height, did he bail out?

1

91 m

2

182 m

3

293 m

4

111 m

Explanation:

supporting img
$h_{1} = 50 m$
$v = \sqrt{2 g h_{1}} = \sqrt{980}$
$\Rightarrow h_{2} = \frac{v^{2} - 9}{2 a} = \frac{980 - 9}{4} = \frac{971}{4}$
$\Rightarrow h_{1} + h_{2} = \frac{1174}{4} \approx 293 m$

AIIMS - 2008

PHXI03:MOTION IN A STRAIGHT LINE

362351 Two balls are dropped from same height at 1 second interval of time. The separation between the two balls after 4 second of the drop of the $1^{st}$ ball is:

1

30 m

2

35 m

3

40 m

4

48 m

Explanation:

Motion of first ball $u = 0, a = g, t = 4 s e c$ . We are consider that the $s_{1}$ is the distance covered by the first ball in 4 seconds.
$s_{1} = u t + \frac{1}{2} g t^{2}$
$0 + \frac{1}{2} \times 10 \times (4)^{2} = 80 m$
Let $s_{2}$ be the distance covered by the second ball in 2 seconds. $t = 3 s e c$ .
$s_{2} = 0 + \frac{1}{2} \times 10 \times (3)^{2} = 45 m$
Separation between the two balls
$S_{1} - S_{2} = 80 m - 45 m = 35 m$

PHXI03:MOTION IN A STRAIGHT LINE

362352 Assertion :
Two balls of different masses are thrown vertically upward with same speed. They will pass through their point of projection in the downward direction with the same speed.
Reason :
The maximum height and downward velocity attained at the point of projection are independent of the mass of ball.

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:

When two objects are thrown upwards with the same initial velocity, they will reach the same maximum height and then fall back down with the same speed.
The maximum height reached by an object depends solely on its initial velocity and the acceleration due to gravity. The downward velocity at the point of projection (when the object starts its descent) is also determined by the initial velocity and gravity, not the mass of the object. $H_{max} = (u^{2} / 2 g)$
Therefore, both the assertion and reason are true, and the reason provides the correct explanation for the assertion. So correct option is (1).

PHXI03:MOTION IN A STRAIGHT LINE

362349 A ball is dropped from a height of 19.6 $m$ . The distance covered by it in the last second is

1 19.6

m

2 14.7

m

3 4.8

m

4 9.8

m

Explanation:

Height, $h = 19.6 m$
Time taken $(t)$ to cover this distance will be given by
$s = u t + \frac{1}{2} a t^{2}$
Here, initial velocity, $u = 0$
$∴ h = \frac{1}{2} g t^{2} \Rightarrow t = \sqrt{\frac{2 h}{g}}$
$\Rightarrow t = \sqrt{\frac{2 \times 19.6}{9.8}} = 2 s$
Velocity acquired in first one second of the motion
$v = u + a t = v = 9.8 \times 1 m / s = 9.8 m / s$
This will be the initial velocity $(u^{'})$ for the last one second of the motion.
$∴$ the distance covered in last second will be
$s = u^{'} t + \frac{1}{2} g + t^{2}$
$\Rightarrow s = 9.8 \times 1 + \frac{1}{2} \times 9.8 \times 1^{2}$
$s = 9.8 + 4.9 = 14.7 m$ .
So correct option is (2)

PHXI03:MOTION IN A STRAIGHT LINE

362350 A parachutist after bailing out falls $50 m$ without friction. When parachute opens, it decelerates at $2 m s^{- 2}$ . He reaches the ground with a speed of $3 m s^{- 1}$ . At what height, did he bail out?

1

91 m

2

182 m

3

293 m

4

111 m

Explanation:

supporting img
$h_{1} = 50 m$
$v = \sqrt{2 g h_{1}} = \sqrt{980}$
$\Rightarrow h_{2} = \frac{v^{2} - 9}{2 a} = \frac{980 - 9}{4} = \frac{971}{4}$
$\Rightarrow h_{1} + h_{2} = \frac{1174}{4} \approx 293 m$

AIIMS - 2008

PHXI03:MOTION IN A STRAIGHT LINE

362351 Two balls are dropped from same height at 1 second interval of time. The separation between the two balls after 4 second of the drop of the $1^{st}$ ball is:

1

30 m

2

35 m

3

40 m

4

48 m

Explanation:

Motion of first ball $u = 0, a = g, t = 4 s e c$ . We are consider that the $s_{1}$ is the distance covered by the first ball in 4 seconds.
$s_{1} = u t + \frac{1}{2} g t^{2}$
$0 + \frac{1}{2} \times 10 \times (4)^{2} = 80 m$
Let $s_{2}$ be the distance covered by the second ball in 2 seconds. $t = 3 s e c$ .
$s_{2} = 0 + \frac{1}{2} \times 10 \times (3)^{2} = 45 m$
Separation between the two balls
$S_{1} - S_{2} = 80 m - 45 m = 35 m$

PHXI03:MOTION IN A STRAIGHT LINE

362352 Assertion :
Two balls of different masses are thrown vertically upward with same speed. They will pass through their point of projection in the downward direction with the same speed.
Reason :
The maximum height and downward velocity attained at the point of projection are independent of the mass of ball.

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:

When two objects are thrown upwards with the same initial velocity, they will reach the same maximum height and then fall back down with the same speed.
The maximum height reached by an object depends solely on its initial velocity and the acceleration due to gravity. The downward velocity at the point of projection (when the object starts its descent) is also determined by the initial velocity and gravity, not the mass of the object. $H_{max} = (u^{2} / 2 g)$
Therefore, both the assertion and reason are true, and the reason provides the correct explanation for the assertion. So correct option is (1).

PHXI03:MOTION IN A STRAIGHT LINE

362349 A ball is dropped from a height of 19.6 $m$ . The distance covered by it in the last second is

1 19.6

m

2 14.7

m

3 4.8

m

4 9.8

m

Explanation:

Height, $h = 19.6 m$
Time taken $(t)$ to cover this distance will be given by
$s = u t + \frac{1}{2} a t^{2}$
Here, initial velocity, $u = 0$
$∴ h = \frac{1}{2} g t^{2} \Rightarrow t = \sqrt{\frac{2 h}{g}}$
$\Rightarrow t = \sqrt{\frac{2 \times 19.6}{9.8}} = 2 s$
Velocity acquired in first one second of the motion
$v = u + a t = v = 9.8 \times 1 m / s = 9.8 m / s$
This will be the initial velocity $(u^{'})$ for the last one second of the motion.
$∴$ the distance covered in last second will be
$s = u^{'} t + \frac{1}{2} g + t^{2}$
$\Rightarrow s = 9.8 \times 1 + \frac{1}{2} \times 9.8 \times 1^{2}$
$s = 9.8 + 4.9 = 14.7 m$ .
So correct option is (2)

PHXI03:MOTION IN A STRAIGHT LINE

362350 A parachutist after bailing out falls $50 m$ without friction. When parachute opens, it decelerates at $2 m s^{- 2}$ . He reaches the ground with a speed of $3 m s^{- 1}$ . At what height, did he bail out?

1

91 m

2

182 m

3

293 m

4

111 m

Explanation:

supporting img
$h_{1} = 50 m$
$v = \sqrt{2 g h_{1}} = \sqrt{980}$
$\Rightarrow h_{2} = \frac{v^{2} - 9}{2 a} = \frac{980 - 9}{4} = \frac{971}{4}$
$\Rightarrow h_{1} + h_{2} = \frac{1174}{4} \approx 293 m$

AIIMS - 2008

PHXI03:MOTION IN A STRAIGHT LINE

362351 Two balls are dropped from same height at 1 second interval of time. The separation between the two balls after 4 second of the drop of the $1^{st}$ ball is:

1

30 m

2

35 m

3

40 m

4

48 m

Explanation:

Motion of first ball $u = 0, a = g, t = 4 s e c$ . We are consider that the $s_{1}$ is the distance covered by the first ball in 4 seconds.
$s_{1} = u t + \frac{1}{2} g t^{2}$
$0 + \frac{1}{2} \times 10 \times (4)^{2} = 80 m$
Let $s_{2}$ be the distance covered by the second ball in 2 seconds. $t = 3 s e c$ .
$s_{2} = 0 + \frac{1}{2} \times 10 \times (3)^{2} = 45 m$
Separation between the two balls
$S_{1} - S_{2} = 80 m - 45 m = 35 m$

PHXI03:MOTION IN A STRAIGHT LINE

362352 Assertion :
Two balls of different masses are thrown vertically upward with same speed. They will pass through their point of projection in the downward direction with the same speed.
Reason :
The maximum height and downward velocity attained at the point of projection are independent of the mass of ball.

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:

When two objects are thrown upwards with the same initial velocity, they will reach the same maximum height and then fall back down with the same speed.
The maximum height reached by an object depends solely on its initial velocity and the acceleration due to gravity. The downward velocity at the point of projection (when the object starts its descent) is also determined by the initial velocity and gravity, not the mass of the object. $H_{max} = (u^{2} / 2 g)$
Therefore, both the assertion and reason are true, and the reason provides the correct explanation for the assertion. So correct option is (1).

PHXI03:MOTION IN A STRAIGHT LINE

362349 A ball is dropped from a height of 19.6 $m$ . The distance covered by it in the last second is

1 19.6

m

2 14.7

m

3 4.8

m

4 9.8

m

Explanation:

Height, $h = 19.6 m$
Time taken $(t)$ to cover this distance will be given by
$s = u t + \frac{1}{2} a t^{2}$
Here, initial velocity, $u = 0$
$∴ h = \frac{1}{2} g t^{2} \Rightarrow t = \sqrt{\frac{2 h}{g}}$
$\Rightarrow t = \sqrt{\frac{2 \times 19.6}{9.8}} = 2 s$
Velocity acquired in first one second of the motion
$v = u + a t = v = 9.8 \times 1 m / s = 9.8 m / s$
This will be the initial velocity $(u^{'})$ for the last one second of the motion.
$∴$ the distance covered in last second will be
$s = u^{'} t + \frac{1}{2} g + t^{2}$
$\Rightarrow s = 9.8 \times 1 + \frac{1}{2} \times 9.8 \times 1^{2}$
$s = 9.8 + 4.9 = 14.7 m$ .
So correct option is (2)

PHXI03:MOTION IN A STRAIGHT LINE

362350 A parachutist after bailing out falls $50 m$ without friction. When parachute opens, it decelerates at $2 m s^{- 2}$ . He reaches the ground with a speed of $3 m s^{- 1}$ . At what height, did he bail out?

1

91 m

2

182 m

3

293 m

4

111 m

Explanation:

supporting img
$h_{1} = 50 m$
$v = \sqrt{2 g h_{1}} = \sqrt{980}$
$\Rightarrow h_{2} = \frac{v^{2} - 9}{2 a} = \frac{980 - 9}{4} = \frac{971}{4}$
$\Rightarrow h_{1} + h_{2} = \frac{1174}{4} \approx 293 m$

AIIMS - 2008

PHXI03:MOTION IN A STRAIGHT LINE

362351 Two balls are dropped from same height at 1 second interval of time. The separation between the two balls after 4 second of the drop of the $1^{st}$ ball is:

1

30 m

2

35 m

3

40 m

4

48 m

Explanation:

Motion of first ball $u = 0, a = g, t = 4 s e c$ . We are consider that the $s_{1}$ is the distance covered by the first ball in 4 seconds.
$s_{1} = u t + \frac{1}{2} g t^{2}$
$0 + \frac{1}{2} \times 10 \times (4)^{2} = 80 m$
Let $s_{2}$ be the distance covered by the second ball in 2 seconds. $t = 3 s e c$ .
$s_{2} = 0 + \frac{1}{2} \times 10 \times (3)^{2} = 45 m$
Separation between the two balls
$S_{1} - S_{2} = 80 m - 45 m = 35 m$

PHXI03:MOTION IN A STRAIGHT LINE

362352 Assertion :
Two balls of different masses are thrown vertically upward with same speed. They will pass through their point of projection in the downward direction with the same speed.
Reason :
The maximum height and downward velocity attained at the point of projection are independent of the mass of ball.

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:

When two objects are thrown upwards with the same initial velocity, they will reach the same maximum height and then fall back down with the same speed.
The maximum height reached by an object depends solely on its initial velocity and the acceleration due to gravity. The downward velocity at the point of projection (when the object starts its descent) is also determined by the initial velocity and gravity, not the mass of the object. $H_{max} = (u^{2} / 2 g)$
Therefore, both the assertion and reason are true, and the reason provides the correct explanation for the assertion. So correct option is (1).

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