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Motion in One Dimensions

141827 A ball is thrown vertically downward with a velocity of $20 m / s$ from the top of a tower. It hits the ground after some time with a velocity of $80 m / s$ . The height of the tower is (g= $10 m / s^{2}$

1

340 m

2

320 m

3

300 m

4

260 m

Explanation:

C Let the height of the tower is $h$ .
The motion of the ball is constant accelerating motion with acceleration $g = 10 m / s^{2}$
$u = 20 m / s$
$v = 80 m / s$
According to equation of motion for constant accelerating motion
(Here downward direction is negative and upward direction is positive)
$v^{2} - u^{2} = 2 (- g) (- h)$
$v^{2} - u^{2} = 2 g h$
$(80)^{2} - (20)^{2} = 2 \times 10 \times h$
$h = \frac{6000}{20}$
$h = 300 m$

NEET (Sep.) 2020

Motion in One Dimensions

141828 A stone thrown vertically upwards with an initial velocity ' $v$ ' from the top of a tower reaches the ground with velocity $3 v$ . The height of tower is

1

3 v^{2} / g

2

4 v^{2} / g

3

6 v^{2} / g

4

9 v^{2} / g

Explanation:

B Given, Initial velocity $= v, u = 3 v$
Now,
$v^{2} = u^{2} - 2 g h$
$v^{2} = (3 v)^{2} - 2 g h$
$v^{2} - 9 v^{2} = - 2 g h$
$2 g h = 8 v^{2}$
$h = \frac{4 v^{2}}{g}$

MHT-CET 2020

Motion in One Dimensions

141829 A stone falls freely such that the distance covered by it in the last second of its motion is equal to the distance covered by it in the first 5 s. It is in air for .......s.

1 12

2 13

3 25

4 26

Explanation:

B Since, stone falls freely, hence its initial velocity, $u = 0$
Distance covered by the stone in first $5 s$ , is given as
$h_{1} = ut + \frac{1}{2} {gt}^{2}$
$= 0 \times 5 + \frac{1}{2} \times 9.8 \times 5^{2}$
$= 122.5$
Since, distance travelled in last second $(t^{th}$ second) is equal to distance travelled in first $5 s$ , hence
$h_{1} = u + \frac{1}{2} g (2 t - 1)$
$122.5 = 0 + \frac{1}{2} \times 9.8 (2 t - 1)$
$2 t - 1 = \frac{122.5 \times 2}{9.8}$
$2 t - 1 = 25 \Rightarrow 2 t = 25 + 1$
$2 t = 26 \Rightarrow t = 13 \sec$

AP EAMCET (21.09.2020) Shift-I

Motion in One Dimensions

141832 A ball is allowed to free fall from the top of a $200 m$ tall tower. Simultaneously another ball is through vertically upwards at a velocity 50 $m / s$ . The balls will meet at a height of

1

121.6 m

from ground

2 125.3 from ground

3

132.1 m

from ground

4 None of above

Explanation:

A original image
The height travelled by dropped ball
$200 - h = \frac{1}{2} {gt}^{2}$
The height travelled by ball thrown vertically upward
$h = 50 t - 0.5 \times g \times t^{2}$
Adding equation (i) and (ii)
$200 = 50 t$
$t = 4 \sec .$
Substituting the value in equation (i)
$200 - h = \frac{1}{2} \times 9.81 \times 4^{2}$
$h = 121.6 m, from the ground$

Assam CEE-2019

Motion in One Dimensions

141827 A ball is thrown vertically downward with a velocity of $20 m / s$ from the top of a tower. It hits the ground after some time with a velocity of $80 m / s$ . The height of the tower is (g= $10 m / s^{2}$

1

340 m

2

320 m

3

300 m

4

260 m

Explanation:

C Let the height of the tower is $h$ .
The motion of the ball is constant accelerating motion with acceleration $g = 10 m / s^{2}$
$u = 20 m / s$
$v = 80 m / s$
According to equation of motion for constant accelerating motion
(Here downward direction is negative and upward direction is positive)
$v^{2} - u^{2} = 2 (- g) (- h)$
$v^{2} - u^{2} = 2 g h$
$(80)^{2} - (20)^{2} = 2 \times 10 \times h$
$h = \frac{6000}{20}$
$h = 300 m$

NEET (Sep.) 2020

Motion in One Dimensions

141828 A stone thrown vertically upwards with an initial velocity ' $v$ ' from the top of a tower reaches the ground with velocity $3 v$ . The height of tower is

1

3 v^{2} / g

2

4 v^{2} / g

3

6 v^{2} / g

4

9 v^{2} / g

Explanation:

B Given, Initial velocity $= v, u = 3 v$
Now,
$v^{2} = u^{2} - 2 g h$
$v^{2} = (3 v)^{2} - 2 g h$
$v^{2} - 9 v^{2} = - 2 g h$
$2 g h = 8 v^{2}$
$h = \frac{4 v^{2}}{g}$

MHT-CET 2020

Motion in One Dimensions

141829 A stone falls freely such that the distance covered by it in the last second of its motion is equal to the distance covered by it in the first 5 s. It is in air for .......s.

1 12

2 13

3 25

4 26

Explanation:

B Since, stone falls freely, hence its initial velocity, $u = 0$
Distance covered by the stone in first $5 s$ , is given as
$h_{1} = ut + \frac{1}{2} {gt}^{2}$
$= 0 \times 5 + \frac{1}{2} \times 9.8 \times 5^{2}$
$= 122.5$
Since, distance travelled in last second $(t^{th}$ second) is equal to distance travelled in first $5 s$ , hence
$h_{1} = u + \frac{1}{2} g (2 t - 1)$
$122.5 = 0 + \frac{1}{2} \times 9.8 (2 t - 1)$
$2 t - 1 = \frac{122.5 \times 2}{9.8}$
$2 t - 1 = 25 \Rightarrow 2 t = 25 + 1$
$2 t = 26 \Rightarrow t = 13 \sec$

AP EAMCET (21.09.2020) Shift-I

Motion in One Dimensions

141832 A ball is allowed to free fall from the top of a $200 m$ tall tower. Simultaneously another ball is through vertically upwards at a velocity 50 $m / s$ . The balls will meet at a height of

1

121.6 m

from ground

2 125.3 from ground

3

132.1 m

from ground

4 None of above

Explanation:

A original image
The height travelled by dropped ball
$200 - h = \frac{1}{2} {gt}^{2}$
The height travelled by ball thrown vertically upward
$h = 50 t - 0.5 \times g \times t^{2}$
Adding equation (i) and (ii)
$200 = 50 t$
$t = 4 \sec .$
Substituting the value in equation (i)
$200 - h = \frac{1}{2} \times 9.81 \times 4^{2}$
$h = 121.6 m, from the ground$

Assam CEE-2019

Motion in One Dimensions

141827 A ball is thrown vertically downward with a velocity of $20 m / s$ from the top of a tower. It hits the ground after some time with a velocity of $80 m / s$ . The height of the tower is (g= $10 m / s^{2}$

1

340 m

2

320 m

3

300 m

4

260 m

Explanation:

C Let the height of the tower is $h$ .
The motion of the ball is constant accelerating motion with acceleration $g = 10 m / s^{2}$
$u = 20 m / s$
$v = 80 m / s$
According to equation of motion for constant accelerating motion
(Here downward direction is negative and upward direction is positive)
$v^{2} - u^{2} = 2 (- g) (- h)$
$v^{2} - u^{2} = 2 g h$
$(80)^{2} - (20)^{2} = 2 \times 10 \times h$
$h = \frac{6000}{20}$
$h = 300 m$

NEET (Sep.) 2020

Motion in One Dimensions

141828 A stone thrown vertically upwards with an initial velocity ' $v$ ' from the top of a tower reaches the ground with velocity $3 v$ . The height of tower is

1

3 v^{2} / g

2

4 v^{2} / g

3

6 v^{2} / g

4

9 v^{2} / g

Explanation:

B Given, Initial velocity $= v, u = 3 v$
Now,
$v^{2} = u^{2} - 2 g h$
$v^{2} = (3 v)^{2} - 2 g h$
$v^{2} - 9 v^{2} = - 2 g h$
$2 g h = 8 v^{2}$
$h = \frac{4 v^{2}}{g}$

MHT-CET 2020

Motion in One Dimensions

141829 A stone falls freely such that the distance covered by it in the last second of its motion is equal to the distance covered by it in the first 5 s. It is in air for .......s.

1 12

2 13

3 25

4 26

Explanation:

B Since, stone falls freely, hence its initial velocity, $u = 0$
Distance covered by the stone in first $5 s$ , is given as
$h_{1} = ut + \frac{1}{2} {gt}^{2}$
$= 0 \times 5 + \frac{1}{2} \times 9.8 \times 5^{2}$
$= 122.5$
Since, distance travelled in last second $(t^{th}$ second) is equal to distance travelled in first $5 s$ , hence
$h_{1} = u + \frac{1}{2} g (2 t - 1)$
$122.5 = 0 + \frac{1}{2} \times 9.8 (2 t - 1)$
$2 t - 1 = \frac{122.5 \times 2}{9.8}$
$2 t - 1 = 25 \Rightarrow 2 t = 25 + 1$
$2 t = 26 \Rightarrow t = 13 \sec$

AP EAMCET (21.09.2020) Shift-I

Motion in One Dimensions

141832 A ball is allowed to free fall from the top of a $200 m$ tall tower. Simultaneously another ball is through vertically upwards at a velocity 50 $m / s$ . The balls will meet at a height of

1

121.6 m

from ground

2 125.3 from ground

3

132.1 m

from ground

4 None of above

Explanation:

A original image
The height travelled by dropped ball
$200 - h = \frac{1}{2} {gt}^{2}$
The height travelled by ball thrown vertically upward
$h = 50 t - 0.5 \times g \times t^{2}$
Adding equation (i) and (ii)
$200 = 50 t$
$t = 4 \sec .$
Substituting the value in equation (i)
$200 - h = \frac{1}{2} \times 9.81 \times 4^{2}$
$h = 121.6 m, from the ground$

Assam CEE-2019

Motion in One Dimensions

141827 A ball is thrown vertically downward with a velocity of $20 m / s$ from the top of a tower. It hits the ground after some time with a velocity of $80 m / s$ . The height of the tower is (g= $10 m / s^{2}$

1

340 m

2

320 m

3

300 m

4

260 m

Explanation:

C Let the height of the tower is $h$ .
The motion of the ball is constant accelerating motion with acceleration $g = 10 m / s^{2}$
$u = 20 m / s$
$v = 80 m / s$
According to equation of motion for constant accelerating motion
(Here downward direction is negative and upward direction is positive)
$v^{2} - u^{2} = 2 (- g) (- h)$
$v^{2} - u^{2} = 2 g h$
$(80)^{2} - (20)^{2} = 2 \times 10 \times h$
$h = \frac{6000}{20}$
$h = 300 m$

NEET (Sep.) 2020

Motion in One Dimensions

141828 A stone thrown vertically upwards with an initial velocity ' $v$ ' from the top of a tower reaches the ground with velocity $3 v$ . The height of tower is

1

3 v^{2} / g

2

4 v^{2} / g

3

6 v^{2} / g

4

9 v^{2} / g

Explanation:

B Given, Initial velocity $= v, u = 3 v$
Now,
$v^{2} = u^{2} - 2 g h$
$v^{2} = (3 v)^{2} - 2 g h$
$v^{2} - 9 v^{2} = - 2 g h$
$2 g h = 8 v^{2}$
$h = \frac{4 v^{2}}{g}$

MHT-CET 2020

Motion in One Dimensions

141829 A stone falls freely such that the distance covered by it in the last second of its motion is equal to the distance covered by it in the first 5 s. It is in air for .......s.

1 12

2 13

3 25

4 26

Explanation:

B Since, stone falls freely, hence its initial velocity, $u = 0$
Distance covered by the stone in first $5 s$ , is given as
$h_{1} = ut + \frac{1}{2} {gt}^{2}$
$= 0 \times 5 + \frac{1}{2} \times 9.8 \times 5^{2}$
$= 122.5$
Since, distance travelled in last second $(t^{th}$ second) is equal to distance travelled in first $5 s$ , hence
$h_{1} = u + \frac{1}{2} g (2 t - 1)$
$122.5 = 0 + \frac{1}{2} \times 9.8 (2 t - 1)$
$2 t - 1 = \frac{122.5 \times 2}{9.8}$
$2 t - 1 = 25 \Rightarrow 2 t = 25 + 1$
$2 t = 26 \Rightarrow t = 13 \sec$

AP EAMCET (21.09.2020) Shift-I

Motion in One Dimensions

141832 A ball is allowed to free fall from the top of a $200 m$ tall tower. Simultaneously another ball is through vertically upwards at a velocity 50 $m / s$ . The balls will meet at a height of

1

121.6 m

from ground

2 125.3 from ground

3

132.1 m

from ground

4 None of above

Explanation:

A original image
The height travelled by dropped ball
$200 - h = \frac{1}{2} {gt}^{2}$
The height travelled by ball thrown vertically upward
$h = 50 t - 0.5 \times g \times t^{2}$
Adding equation (i) and (ii)
$200 = 50 t$
$t = 4 \sec .$
Substituting the value in equation (i)
$200 - h = \frac{1}{2} \times 9.81 \times 4^{2}$
$h = 121.6 m, from the ground$

Assam CEE-2019

Question Bank Series

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