QBManager

Motion in One Dimensions

141798 A ball is thrown upward from the top of a building at an angle of $30^{\circ}$ to the horizontal and with an initial speed of $20 m s^{- 1}$ . If the ball strikes the ground after $3 s$ , then the height of the buildings is-
(acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

10 m

2

15 m

3

20 m

4

25 m

Explanation:

B Given data,
$u = 20 m / s, θ = 30^{\circ}$
$t = 3 \sec$
original image
So, according to question,
When the ball is thrown upward
$s = ut - \frac{1}{2} {gt}^{2}$
$s = 20 \sin 30^{\circ} \times 3 - \frac{1}{2} \times 10 \times 3^{2}$
$s = 30 - 45$
$∴ s = - 15 m \Rightarrow - + - s = 15 m$

AP EAMCET-07.07.2022

Motion in One Dimensions

141799 When a ball is dropped from a height $h$ it takes $t$ sec to reach the ground. If the same experiment is done on a different planet having the mass 100 times the earth's mass and radius 10 times the earth's radius, then the time it will take to cover the same height in the new planet is

1

ts

2 100 ts

3

\frac{t}{100} s

4

\frac{t}{10} s

Explanation:

A If a ball is dropped from a height $h$ it takes $t$ sec to reach the ground
So, $u = 0$ , distance travelled $h h$
$∵ s = ut + \frac{1}{2} a t^{2}$
Put the given value,
$h = 0 + \frac{1}{2} {gt}^{2} (∵ a = g at earth)$
$h = \frac{1}{2} {gt}^{2}$
Now for different planet having,
$M = 100 Me (given)$
$R = 10 Re$
So to calculate gravitational acceleration on that planet, We know
$g = \frac{GM}{R^{2}}$
$∴ g^{'} = \frac{G (100 Me)}{(10 Re)^{2}}$
$g^{'} = \frac{100 {GM}_{e}}{100 R_{e}^{2}} = \frac{GM}{R^{2}} = g$
We can say that $g$ for another planet is as same as $g$ . Hence, if ball takes $t^{'}$ second on other planet
$s = {ut}^{'} + \frac{1}{2} g^{' t^{' 2}} \Rightarrow h^{'} = 0 + \frac{1}{2} g^{'} t^{' 2}$
To cover same distance means
$h = h^{'}$
Hence,
$\frac{1}{2} {gt}^{2} = \frac{1}{2} {gt}^{' 2} \Rightarrow t^{'} = tsec$

AP EAMCET-05.07.2022

Motion in One Dimensions

141800 An object of mass $5 kg$ is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of $10 N$ throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use $g = 10 {ms}^{- 2}$ ]

1

1 : 1

2

\sqrt{2} : \sqrt{3}

3

\sqrt{3} : \sqrt{2}

4

2 : 3

Explanation:

B For upward motion acceleration $= g + a$ for downward motion acceleration $= g - a$
$h = ut + \frac{1}{2} {at}_{a}^{2}$
$= 0 + \frac{1}{2} (g + a) t_{a}^{2}$
Then $h = 0 + \frac{1}{2} (g - a) t_{d}^{2}$
$f = m a$
$a = f / m = \frac{10}{5} = 2 m / s^{2}$
Equation (i) divided by equation (ii)
$\frac{t_{a}}{t_{d}} = \sqrt{\frac{g - a}{g + a}} \Rightarrow \frac{t_{a}}{t_{d}} = \sqrt{\frac{10 - 2}{10 + 2}} = \sqrt{\frac{8}{12}}$
$\frac{t_{a}}{t_{d}} = \sqrt{\frac{2}{3}}$

JEE Main-24.06.2022

Motion in One Dimensions

141801 A ball is thrown upward from the top of a building at angle of $30^{\circ}$ to the horizontal with an initial speed of $15 {ms}^{- 1}$ . If the ball hits the ground after $3 s$ , then the height of building is(acceleration due to gravity $= 10 {ms}^{- 2}$ )

1

30 m

2

12.5 m

3

25.5 m

4

22.5 m

Explanation:

D We know
original image
Then, from second equation of motion
$| h | = | u_{1} t + \frac{1}{2} {at}^{2} |$
$| h | = | \frac{15}{2} \times 3 - \frac{1}{2} \times 10 \times (3)^{2} |$
$| h | = | - 22.5 |$
$h = 22.5 m$

AP EAMCET-08.07.2022

Motion in One Dimensions

141802 A hot air balloon is descending with a constant acceleration of $2 {ms}^{- 2}$ . The mass of the balloon and its contents is $600 kg$ . Assuming the volume of balloon is kept constant the mass of the balloon required so that the balloon starts ascending with the same acceleration.
(Acceleration due to gravity $= 10 m s^{- 2}$ )

1

400 kg

2

200 kg

3

100 kg

4

500 kg

Explanation:

A Let $m kg$ mass is removed
original image
In case-I -
The equation of motion of the balloon.
$Mg - F = Ma$
In case - II -
The equation of motion of the balloon
$F - (M - m) g = (M - m) a$
$F - Mg + mg = Ma - ma$
Adding equation (i) and (ii) we get
$mg = 2 Ma - ma$
$m (g + a) = 2 Ma$
$m = \frac{2 Ma}{(g + a)} \Rightarrow m = \frac{2 \times 600 \times 2}{(10 + 2)}$
$= 200 kg$
Thus, required mass $= 600 - 200$
$= 400 kg$

AP EAMCET-12.07.2022

Motion in One Dimensions

141798 A ball is thrown upward from the top of a building at an angle of $30^{\circ}$ to the horizontal and with an initial speed of $20 m s^{- 1}$ . If the ball strikes the ground after $3 s$ , then the height of the buildings is-
(acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

10 m

2

15 m

3

20 m

4

25 m

Explanation:

B Given data,
$u = 20 m / s, θ = 30^{\circ}$
$t = 3 \sec$
original image
So, according to question,
When the ball is thrown upward
$s = ut - \frac{1}{2} {gt}^{2}$
$s = 20 \sin 30^{\circ} \times 3 - \frac{1}{2} \times 10 \times 3^{2}$
$s = 30 - 45$
$∴ s = - 15 m \Rightarrow - + - s = 15 m$

AP EAMCET-07.07.2022

Motion in One Dimensions

141799 When a ball is dropped from a height $h$ it takes $t$ sec to reach the ground. If the same experiment is done on a different planet having the mass 100 times the earth's mass and radius 10 times the earth's radius, then the time it will take to cover the same height in the new planet is

1

ts

2 100 ts

3

\frac{t}{100} s

4

\frac{t}{10} s

Explanation:

A If a ball is dropped from a height $h$ it takes $t$ sec to reach the ground
So, $u = 0$ , distance travelled $h h$
$∵ s = ut + \frac{1}{2} a t^{2}$
Put the given value,
$h = 0 + \frac{1}{2} {gt}^{2} (∵ a = g at earth)$
$h = \frac{1}{2} {gt}^{2}$
Now for different planet having,
$M = 100 Me (given)$
$R = 10 Re$
So to calculate gravitational acceleration on that planet, We know
$g = \frac{GM}{R^{2}}$
$∴ g^{'} = \frac{G (100 Me)}{(10 Re)^{2}}$
$g^{'} = \frac{100 {GM}_{e}}{100 R_{e}^{2}} = \frac{GM}{R^{2}} = g$
We can say that $g$ for another planet is as same as $g$ . Hence, if ball takes $t^{'}$ second on other planet
$s = {ut}^{'} + \frac{1}{2} g^{' t^{' 2}} \Rightarrow h^{'} = 0 + \frac{1}{2} g^{'} t^{' 2}$
To cover same distance means
$h = h^{'}$
Hence,
$\frac{1}{2} {gt}^{2} = \frac{1}{2} {gt}^{' 2} \Rightarrow t^{'} = tsec$

AP EAMCET-05.07.2022

Motion in One Dimensions

141800 An object of mass $5 kg$ is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of $10 N$ throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use $g = 10 {ms}^{- 2}$ ]

1

1 : 1

2

\sqrt{2} : \sqrt{3}

3

\sqrt{3} : \sqrt{2}

4

2 : 3

Explanation:

B For upward motion acceleration $= g + a$ for downward motion acceleration $= g - a$
$h = ut + \frac{1}{2} {at}_{a}^{2}$
$= 0 + \frac{1}{2} (g + a) t_{a}^{2}$
Then $h = 0 + \frac{1}{2} (g - a) t_{d}^{2}$
$f = m a$
$a = f / m = \frac{10}{5} = 2 m / s^{2}$
Equation (i) divided by equation (ii)
$\frac{t_{a}}{t_{d}} = \sqrt{\frac{g - a}{g + a}} \Rightarrow \frac{t_{a}}{t_{d}} = \sqrt{\frac{10 - 2}{10 + 2}} = \sqrt{\frac{8}{12}}$
$\frac{t_{a}}{t_{d}} = \sqrt{\frac{2}{3}}$

JEE Main-24.06.2022

Motion in One Dimensions

141801 A ball is thrown upward from the top of a building at angle of $30^{\circ}$ to the horizontal with an initial speed of $15 {ms}^{- 1}$ . If the ball hits the ground after $3 s$ , then the height of building is(acceleration due to gravity $= 10 {ms}^{- 2}$ )

1

30 m

2

12.5 m

3

25.5 m

4

22.5 m

Explanation:

D We know
original image
Then, from second equation of motion
$| h | = | u_{1} t + \frac{1}{2} {at}^{2} |$
$| h | = | \frac{15}{2} \times 3 - \frac{1}{2} \times 10 \times (3)^{2} |$
$| h | = | - 22.5 |$
$h = 22.5 m$

AP EAMCET-08.07.2022

Motion in One Dimensions

141802 A hot air balloon is descending with a constant acceleration of $2 {ms}^{- 2}$ . The mass of the balloon and its contents is $600 kg$ . Assuming the volume of balloon is kept constant the mass of the balloon required so that the balloon starts ascending with the same acceleration.
(Acceleration due to gravity $= 10 m s^{- 2}$ )

1

400 kg

2

200 kg

3

100 kg

4

500 kg

Explanation:

A Let $m kg$ mass is removed
original image
In case-I -
The equation of motion of the balloon.
$Mg - F = Ma$
In case - II -
The equation of motion of the balloon
$F - (M - m) g = (M - m) a$
$F - Mg + mg = Ma - ma$
Adding equation (i) and (ii) we get
$mg = 2 Ma - ma$
$m (g + a) = 2 Ma$
$m = \frac{2 Ma}{(g + a)} \Rightarrow m = \frac{2 \times 600 \times 2}{(10 + 2)}$
$= 200 kg$
Thus, required mass $= 600 - 200$
$= 400 kg$

AP EAMCET-12.07.2022

Motion in One Dimensions

141798 A ball is thrown upward from the top of a building at an angle of $30^{\circ}$ to the horizontal and with an initial speed of $20 m s^{- 1}$ . If the ball strikes the ground after $3 s$ , then the height of the buildings is-
(acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

10 m

2

15 m

3

20 m

4

25 m

Explanation:

B Given data,
$u = 20 m / s, θ = 30^{\circ}$
$t = 3 \sec$
original image
So, according to question,
When the ball is thrown upward
$s = ut - \frac{1}{2} {gt}^{2}$
$s = 20 \sin 30^{\circ} \times 3 - \frac{1}{2} \times 10 \times 3^{2}$
$s = 30 - 45$
$∴ s = - 15 m \Rightarrow - + - s = 15 m$

AP EAMCET-07.07.2022

Motion in One Dimensions

141799 When a ball is dropped from a height $h$ it takes $t$ sec to reach the ground. If the same experiment is done on a different planet having the mass 100 times the earth's mass and radius 10 times the earth's radius, then the time it will take to cover the same height in the new planet is

1

ts

2 100 ts

3

\frac{t}{100} s

4

\frac{t}{10} s

Explanation:

A If a ball is dropped from a height $h$ it takes $t$ sec to reach the ground
So, $u = 0$ , distance travelled $h h$
$∵ s = ut + \frac{1}{2} a t^{2}$
Put the given value,
$h = 0 + \frac{1}{2} {gt}^{2} (∵ a = g at earth)$
$h = \frac{1}{2} {gt}^{2}$
Now for different planet having,
$M = 100 Me (given)$
$R = 10 Re$
So to calculate gravitational acceleration on that planet, We know
$g = \frac{GM}{R^{2}}$
$∴ g^{'} = \frac{G (100 Me)}{(10 Re)^{2}}$
$g^{'} = \frac{100 {GM}_{e}}{100 R_{e}^{2}} = \frac{GM}{R^{2}} = g$
We can say that $g$ for another planet is as same as $g$ . Hence, if ball takes $t^{'}$ second on other planet
$s = {ut}^{'} + \frac{1}{2} g^{' t^{' 2}} \Rightarrow h^{'} = 0 + \frac{1}{2} g^{'} t^{' 2}$
To cover same distance means
$h = h^{'}$
Hence,
$\frac{1}{2} {gt}^{2} = \frac{1}{2} {gt}^{' 2} \Rightarrow t^{'} = tsec$

AP EAMCET-05.07.2022

Motion in One Dimensions

141800 An object of mass $5 kg$ is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of $10 N$ throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use $g = 10 {ms}^{- 2}$ ]

1

1 : 1

2

\sqrt{2} : \sqrt{3}

3

\sqrt{3} : \sqrt{2}

4

2 : 3

Explanation:

B For upward motion acceleration $= g + a$ for downward motion acceleration $= g - a$
$h = ut + \frac{1}{2} {at}_{a}^{2}$
$= 0 + \frac{1}{2} (g + a) t_{a}^{2}$
Then $h = 0 + \frac{1}{2} (g - a) t_{d}^{2}$
$f = m a$
$a = f / m = \frac{10}{5} = 2 m / s^{2}$
Equation (i) divided by equation (ii)
$\frac{t_{a}}{t_{d}} = \sqrt{\frac{g - a}{g + a}} \Rightarrow \frac{t_{a}}{t_{d}} = \sqrt{\frac{10 - 2}{10 + 2}} = \sqrt{\frac{8}{12}}$
$\frac{t_{a}}{t_{d}} = \sqrt{\frac{2}{3}}$

JEE Main-24.06.2022

Motion in One Dimensions

141801 A ball is thrown upward from the top of a building at angle of $30^{\circ}$ to the horizontal with an initial speed of $15 {ms}^{- 1}$ . If the ball hits the ground after $3 s$ , then the height of building is(acceleration due to gravity $= 10 {ms}^{- 2}$ )

1

30 m

2

12.5 m

3

25.5 m

4

22.5 m

Explanation:

D We know
original image
Then, from second equation of motion
$| h | = | u_{1} t + \frac{1}{2} {at}^{2} |$
$| h | = | \frac{15}{2} \times 3 - \frac{1}{2} \times 10 \times (3)^{2} |$
$| h | = | - 22.5 |$
$h = 22.5 m$

AP EAMCET-08.07.2022

Motion in One Dimensions

141802 A hot air balloon is descending with a constant acceleration of $2 {ms}^{- 2}$ . The mass of the balloon and its contents is $600 kg$ . Assuming the volume of balloon is kept constant the mass of the balloon required so that the balloon starts ascending with the same acceleration.
(Acceleration due to gravity $= 10 m s^{- 2}$ )

1

400 kg

2

200 kg

3

100 kg

4

500 kg

Explanation:

A Let $m kg$ mass is removed
original image
In case-I -
The equation of motion of the balloon.
$Mg - F = Ma$
In case - II -
The equation of motion of the balloon
$F - (M - m) g = (M - m) a$
$F - Mg + mg = Ma - ma$
Adding equation (i) and (ii) we get
$mg = 2 Ma - ma$
$m (g + a) = 2 Ma$
$m = \frac{2 Ma}{(g + a)} \Rightarrow m = \frac{2 \times 600 \times 2}{(10 + 2)}$
$= 200 kg$
Thus, required mass $= 600 - 200$
$= 400 kg$

AP EAMCET-12.07.2022

Motion in One Dimensions

141798 A ball is thrown upward from the top of a building at an angle of $30^{\circ}$ to the horizontal and with an initial speed of $20 m s^{- 1}$ . If the ball strikes the ground after $3 s$ , then the height of the buildings is-
(acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

10 m

2

15 m

3

20 m

4

25 m

Explanation:

B Given data,
$u = 20 m / s, θ = 30^{\circ}$
$t = 3 \sec$
original image
So, according to question,
When the ball is thrown upward
$s = ut - \frac{1}{2} {gt}^{2}$
$s = 20 \sin 30^{\circ} \times 3 - \frac{1}{2} \times 10 \times 3^{2}$
$s = 30 - 45$
$∴ s = - 15 m \Rightarrow - + - s = 15 m$

AP EAMCET-07.07.2022

Motion in One Dimensions

141799 When a ball is dropped from a height $h$ it takes $t$ sec to reach the ground. If the same experiment is done on a different planet having the mass 100 times the earth's mass and radius 10 times the earth's radius, then the time it will take to cover the same height in the new planet is

1

ts

2 100 ts

3

\frac{t}{100} s

4

\frac{t}{10} s

Explanation:

A If a ball is dropped from a height $h$ it takes $t$ sec to reach the ground
So, $u = 0$ , distance travelled $h h$
$∵ s = ut + \frac{1}{2} a t^{2}$
Put the given value,
$h = 0 + \frac{1}{2} {gt}^{2} (∵ a = g at earth)$
$h = \frac{1}{2} {gt}^{2}$
Now for different planet having,
$M = 100 Me (given)$
$R = 10 Re$
So to calculate gravitational acceleration on that planet, We know
$g = \frac{GM}{R^{2}}$
$∴ g^{'} = \frac{G (100 Me)}{(10 Re)^{2}}$
$g^{'} = \frac{100 {GM}_{e}}{100 R_{e}^{2}} = \frac{GM}{R^{2}} = g$
We can say that $g$ for another planet is as same as $g$ . Hence, if ball takes $t^{'}$ second on other planet
$s = {ut}^{'} + \frac{1}{2} g^{' t^{' 2}} \Rightarrow h^{'} = 0 + \frac{1}{2} g^{'} t^{' 2}$
To cover same distance means
$h = h^{'}$
Hence,
$\frac{1}{2} {gt}^{2} = \frac{1}{2} {gt}^{' 2} \Rightarrow t^{'} = tsec$

AP EAMCET-05.07.2022

Motion in One Dimensions

141800 An object of mass $5 kg$ is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of $10 N$ throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use $g = 10 {ms}^{- 2}$ ]

1

1 : 1

2

\sqrt{2} : \sqrt{3}

3

\sqrt{3} : \sqrt{2}

4

2 : 3

Explanation:

B For upward motion acceleration $= g + a$ for downward motion acceleration $= g - a$
$h = ut + \frac{1}{2} {at}_{a}^{2}$
$= 0 + \frac{1}{2} (g + a) t_{a}^{2}$
Then $h = 0 + \frac{1}{2} (g - a) t_{d}^{2}$
$f = m a$
$a = f / m = \frac{10}{5} = 2 m / s^{2}$
Equation (i) divided by equation (ii)
$\frac{t_{a}}{t_{d}} = \sqrt{\frac{g - a}{g + a}} \Rightarrow \frac{t_{a}}{t_{d}} = \sqrt{\frac{10 - 2}{10 + 2}} = \sqrt{\frac{8}{12}}$
$\frac{t_{a}}{t_{d}} = \sqrt{\frac{2}{3}}$

JEE Main-24.06.2022

Motion in One Dimensions

141801 A ball is thrown upward from the top of a building at angle of $30^{\circ}$ to the horizontal with an initial speed of $15 {ms}^{- 1}$ . If the ball hits the ground after $3 s$ , then the height of building is(acceleration due to gravity $= 10 {ms}^{- 2}$ )

1

30 m

2

12.5 m

3

25.5 m

4

22.5 m

Explanation:

D We know
original image
Then, from second equation of motion
$| h | = | u_{1} t + \frac{1}{2} {at}^{2} |$
$| h | = | \frac{15}{2} \times 3 - \frac{1}{2} \times 10 \times (3)^{2} |$
$| h | = | - 22.5 |$
$h = 22.5 m$

AP EAMCET-08.07.2022

Motion in One Dimensions

141802 A hot air balloon is descending with a constant acceleration of $2 {ms}^{- 2}$ . The mass of the balloon and its contents is $600 kg$ . Assuming the volume of balloon is kept constant the mass of the balloon required so that the balloon starts ascending with the same acceleration.
(Acceleration due to gravity $= 10 m s^{- 2}$ )

1

400 kg

2

200 kg

3

100 kg

4

500 kg

Explanation:

A Let $m kg$ mass is removed
original image
In case-I -
The equation of motion of the balloon.
$Mg - F = Ma$
In case - II -
The equation of motion of the balloon
$F - (M - m) g = (M - m) a$
$F - Mg + mg = Ma - ma$
Adding equation (i) and (ii) we get
$mg = 2 Ma - ma$
$m (g + a) = 2 Ma$
$m = \frac{2 Ma}{(g + a)} \Rightarrow m = \frac{2 \times 600 \times 2}{(10 + 2)}$
$= 200 kg$
Thus, required mass $= 600 - 200$
$= 400 kg$

AP EAMCET-12.07.2022

Motion in One Dimensions

141798 A ball is thrown upward from the top of a building at an angle of $30^{\circ}$ to the horizontal and with an initial speed of $20 m s^{- 1}$ . If the ball strikes the ground after $3 s$ , then the height of the buildings is-
(acceleration due to gravity $= 10 {m s}^{- 2}$ )

1

10 m

2

15 m

3

20 m

4

25 m

Explanation:

B Given data,
$u = 20 m / s, θ = 30^{\circ}$
$t = 3 \sec$
original image
So, according to question,
When the ball is thrown upward
$s = ut - \frac{1}{2} {gt}^{2}$
$s = 20 \sin 30^{\circ} \times 3 - \frac{1}{2} \times 10 \times 3^{2}$
$s = 30 - 45$
$∴ s = - 15 m \Rightarrow - + - s = 15 m$

AP EAMCET-07.07.2022

Motion in One Dimensions

141799 When a ball is dropped from a height $h$ it takes $t$ sec to reach the ground. If the same experiment is done on a different planet having the mass 100 times the earth's mass and radius 10 times the earth's radius, then the time it will take to cover the same height in the new planet is

1

ts

2 100 ts

3

\frac{t}{100} s

4

\frac{t}{10} s

Explanation:

A If a ball is dropped from a height $h$ it takes $t$ sec to reach the ground
So, $u = 0$ , distance travelled $h h$
$∵ s = ut + \frac{1}{2} a t^{2}$
Put the given value,
$h = 0 + \frac{1}{2} {gt}^{2} (∵ a = g at earth)$
$h = \frac{1}{2} {gt}^{2}$
Now for different planet having,
$M = 100 Me (given)$
$R = 10 Re$
So to calculate gravitational acceleration on that planet, We know
$g = \frac{GM}{R^{2}}$
$∴ g^{'} = \frac{G (100 Me)}{(10 Re)^{2}}$
$g^{'} = \frac{100 {GM}_{e}}{100 R_{e}^{2}} = \frac{GM}{R^{2}} = g$
We can say that $g$ for another planet is as same as $g$ . Hence, if ball takes $t^{'}$ second on other planet
$s = {ut}^{'} + \frac{1}{2} g^{' t^{' 2}} \Rightarrow h^{'} = 0 + \frac{1}{2} g^{'} t^{' 2}$
To cover same distance means
$h = h^{'}$
Hence,
$\frac{1}{2} {gt}^{2} = \frac{1}{2} {gt}^{' 2} \Rightarrow t^{'} = tsec$

AP EAMCET-05.07.2022

Motion in One Dimensions

141800 An object of mass $5 kg$ is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of $10 N$ throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use $g = 10 {ms}^{- 2}$ ]

1

1 : 1

2

\sqrt{2} : \sqrt{3}

3

\sqrt{3} : \sqrt{2}

4

2 : 3

Explanation:

B For upward motion acceleration $= g + a$ for downward motion acceleration $= g - a$
$h = ut + \frac{1}{2} {at}_{a}^{2}$
$= 0 + \frac{1}{2} (g + a) t_{a}^{2}$
Then $h = 0 + \frac{1}{2} (g - a) t_{d}^{2}$
$f = m a$
$a = f / m = \frac{10}{5} = 2 m / s^{2}$
Equation (i) divided by equation (ii)
$\frac{t_{a}}{t_{d}} = \sqrt{\frac{g - a}{g + a}} \Rightarrow \frac{t_{a}}{t_{d}} = \sqrt{\frac{10 - 2}{10 + 2}} = \sqrt{\frac{8}{12}}$
$\frac{t_{a}}{t_{d}} = \sqrt{\frac{2}{3}}$

JEE Main-24.06.2022

Motion in One Dimensions

141801 A ball is thrown upward from the top of a building at angle of $30^{\circ}$ to the horizontal with an initial speed of $15 {ms}^{- 1}$ . If the ball hits the ground after $3 s$ , then the height of building is(acceleration due to gravity $= 10 {ms}^{- 2}$ )

1

30 m

2

12.5 m

3

25.5 m

4

22.5 m

Explanation:

D We know
original image
Then, from second equation of motion
$| h | = | u_{1} t + \frac{1}{2} {at}^{2} |$
$| h | = | \frac{15}{2} \times 3 - \frac{1}{2} \times 10 \times (3)^{2} |$
$| h | = | - 22.5 |$
$h = 22.5 m$

AP EAMCET-08.07.2022

Motion in One Dimensions

141802 A hot air balloon is descending with a constant acceleration of $2 {ms}^{- 2}$ . The mass of the balloon and its contents is $600 kg$ . Assuming the volume of balloon is kept constant the mass of the balloon required so that the balloon starts ascending with the same acceleration.
(Acceleration due to gravity $= 10 m s^{- 2}$ )

1

400 kg

2

200 kg

3

100 kg

4

500 kg

Explanation:

A Let $m kg$ mass is removed
original image
In case-I -
The equation of motion of the balloon.
$Mg - F = Ma$
In case - II -
The equation of motion of the balloon
$F - (M - m) g = (M - m) a$
$F - Mg + mg = Ma - ma$
Adding equation (i) and (ii) we get
$mg = 2 Ma - ma$
$m (g + a) = 2 Ma$
$m = \frac{2 Ma}{(g + a)} \Rightarrow m = \frac{2 \times 600 \times 2}{(10 + 2)}$
$= 200 kg$
Thus, required mass $= 600 - 200$
$= 400 kg$

AP EAMCET-12.07.2022