QBManager

Motion in One Dimensions

141785 A ball when thrown upward from a tower with a given velocity takes time $t_{1}$ to reach the ground. The same ball when thrown downward with same velocity from the tower takes time $t_{2}$ to reach the ground. If the ball is dropped freely from the same tower, the time taken to reach the ground will be

1

t_{1} - t_{2}

2

\frac{t_{1} + t_{2}}{2}

3

\sqrt{t_{1} \times t_{2}}

4

\sqrt{t_{1}^{2} - t_{2}^{2}}

Explanation:

C For case $I^{st}$ ,
original image
$- h = {vt}_{1} - \frac{1}{2} {gt}_{1}^{2}$
$+ \frac{h}{t_{1}} = - v + \frac{1}{2} {gt}_{1}$
For case II $^{nd}$ ,
$- h = - {vt}_{2} - \frac{1}{2} {gt}_{2}^{2}$
$\frac{h}{t_{2}} = v + \frac{1}{2} {gt}_{2} \dots .$
Adding equation (i) and equation (ii), we get
$\frac{h}{t_{1}} + \frac{h}{t_{2}} = \frac{1}{2} g t_{1} + \frac{1}{2} g t_{2}$
$h (\frac{1}{t_{1}} + \frac{1}{t_{2}}) = \frac{g}{2} (t_{1} + t_{2}) \Rightarrow h \frac{(t_{1} + t_{2})}{t_{1} \cdot t_{2}} = \frac{g}{2} (t_{1} + t_{2})$
$h = \frac{g t_{1} t_{2}}{2} \dots . . . (iii)$
For case III $^{rd}$ ,
$u = 0, t = ?$
$- h = 0 \times t - \frac{1}{2} {gt}^{2} \Rightarrow h = \frac{1}{2} {gt}^{2}$
$t^{2} = \frac{2}{g} \times h$
Putting the value of $h$ from equation (iii) in equation (iv) we get-
$t^{2} = \frac{2}{g} \times \frac{g t_{1} \cdot t_{2}}{2} \Rightarrow t = \sqrt{t_{1} \cdot t_{2}}$

SCRA-2015

Motion in One Dimensions

141786 A body dropped from top of a tower fall through $60 m$ during the last two seconds of its fall. The height of tower is : $(g = 10 m / s^{2})$

1

95 m

2

60 m

3

80 m

4

90 m

Explanation:

C Let the total time be $t$
original image
Initial velocity $(u) = 0$
After $(t - 2)$ sec, final velocity
$v = 0 + g \times (t - 2)$
$v = g (t - 2)$
Now $60 m$ height covered with velocity $v$ in $2 \sec -$
$∴ 60 = g (t - 2) \times 2 + \frac{1}{2} \times g (2)^{2}$
$60 = 20 (t - 2) + \frac{1}{2} \times 10 \times 4$
$t = 4 \sec$
Hence, total height $= \frac{1}{2} {gt}^{2}$
$= \frac{1}{2} \times 10 \times 16$
$= 80 m$

UPSEE - 2005

Motion in One Dimensions

141787 The ratio of distance traversed in successive intervals of time when a body falls freely under gravity from certain height is

1

1 : 2 : 3

2

1 : 5 : 9

3

1 : 3 : 5

4

\sqrt{1} : \sqrt{2} : \sqrt{3}

5

1 : 4 : 9

Explanation:

C (i) Distance travelled in $1 \sec$ ,
$s_{1} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times (1)^{2}$
$s_{1} = 5 m$
(ii) Distance travelled in $2 \sec$
$s_{2} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times (2)^{2}$ .
$s_{2} = 20 m$
(iii) Distance travelled in $3 \sec$
$s_{3} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times (3)^{2}$
$s_{3} = 45 m$
The ratio of distance,
$s_{1} : s_{2} : s_{3} = 5 : (20 - 5) : (45 - 20)$
$= 5 : 15 : 25$
$= 1 : 3 : 5$

Kerala CEE- 2014

Motion in One Dimensions

141788 A stone is thrown vertically upwards. When the stone is at a height equal to half of its maximum height its speed will be $10 m / s$ , then the maximum height attained by the stone is (Take $g = 10 m / s^{2}$ )

1

3 m

2

15 m

3

1 m

4

10 m

Explanation:

D When the stone is at a height equal to half of its maximum height
From, $v^{2} = u^{2} - 2 g h$
$v = 10 m / s$
$(10)^{2} = u^{2} - 2 \times 10 \times \frac{h}{2}$
$100 = u^{2} - 10 h$
Speed at the maximum height,
$v = 0$
$v^{2} = u^{2} - 2 g h$
$0 = u^{2} - 2 \times 10 \times h$
$u^{2} = 20 h \dots \dots \dots \dots . .$
From equation (i) and (ii), we get-
$100 = 20 h - 10 h$
$10 h = 100 \Rightarrow h = 10 m$

JCECE-2014

Motion in One Dimensions

141789 A ball is released from the top of a tower of height $h$ meters. It takes $T$ seconds to reach the ground. What is the position of the ball at $\frac{T}{3}$ second

1

\frac{8 h}{9}

meters from the ground

2

\frac{7 h}{9}

meters from the ground

3

\frac{h}{9}

meters from the ground

4

\frac{17 h}{18}

meters from the ground

Explanation:

A Given,
Height of tower $= h$
Time taken to reach the ground $= T$
From,
$h = ut + \frac{1}{2} {gt}^{2}$
$h = 0 + \frac{1}{2} {gT}^{2}$
Distance travelled in $T / 3$ second
$h = \frac{1}{2} g {(\frac{T}{3})}^{2} = \frac{1}{2} g \frac{T^{2}}{9}$
$= \frac{h}{9} [from eq$
$(i)]$
$∴$ Distance of ball from the ground $= h - \frac{h}{9} = \frac{8 h}{9}$

AIIMS-2012

Motion in One Dimensions

141785 A ball when thrown upward from a tower with a given velocity takes time $t_{1}$ to reach the ground. The same ball when thrown downward with same velocity from the tower takes time $t_{2}$ to reach the ground. If the ball is dropped freely from the same tower, the time taken to reach the ground will be

1

t_{1} - t_{2}

2

\frac{t_{1} + t_{2}}{2}

3

\sqrt{t_{1} \times t_{2}}

4

\sqrt{t_{1}^{2} - t_{2}^{2}}

Explanation:

C For case $I^{st}$ ,
original image
$- h = {vt}_{1} - \frac{1}{2} {gt}_{1}^{2}$
$+ \frac{h}{t_{1}} = - v + \frac{1}{2} {gt}_{1}$
For case II $^{nd}$ ,
$- h = - {vt}_{2} - \frac{1}{2} {gt}_{2}^{2}$
$\frac{h}{t_{2}} = v + \frac{1}{2} {gt}_{2} \dots .$
Adding equation (i) and equation (ii), we get
$\frac{h}{t_{1}} + \frac{h}{t_{2}} = \frac{1}{2} g t_{1} + \frac{1}{2} g t_{2}$
$h (\frac{1}{t_{1}} + \frac{1}{t_{2}}) = \frac{g}{2} (t_{1} + t_{2}) \Rightarrow h \frac{(t_{1} + t_{2})}{t_{1} \cdot t_{2}} = \frac{g}{2} (t_{1} + t_{2})$
$h = \frac{g t_{1} t_{2}}{2} \dots . . . (iii)$
For case III $^{rd}$ ,
$u = 0, t = ?$
$- h = 0 \times t - \frac{1}{2} {gt}^{2} \Rightarrow h = \frac{1}{2} {gt}^{2}$
$t^{2} = \frac{2}{g} \times h$
Putting the value of $h$ from equation (iii) in equation (iv) we get-
$t^{2} = \frac{2}{g} \times \frac{g t_{1} \cdot t_{2}}{2} \Rightarrow t = \sqrt{t_{1} \cdot t_{2}}$

SCRA-2015

Motion in One Dimensions

141786 A body dropped from top of a tower fall through $60 m$ during the last two seconds of its fall. The height of tower is : $(g = 10 m / s^{2})$

1

95 m

2

60 m

3

80 m

4

90 m

Explanation:

C Let the total time be $t$
original image
Initial velocity $(u) = 0$
After $(t - 2)$ sec, final velocity
$v = 0 + g \times (t - 2)$
$v = g (t - 2)$
Now $60 m$ height covered with velocity $v$ in $2 \sec -$
$∴ 60 = g (t - 2) \times 2 + \frac{1}{2} \times g (2)^{2}$
$60 = 20 (t - 2) + \frac{1}{2} \times 10 \times 4$
$t = 4 \sec$
Hence, total height $= \frac{1}{2} {gt}^{2}$
$= \frac{1}{2} \times 10 \times 16$
$= 80 m$

UPSEE - 2005

Motion in One Dimensions

141787 The ratio of distance traversed in successive intervals of time when a body falls freely under gravity from certain height is

1

1 : 2 : 3

2

1 : 5 : 9

3

1 : 3 : 5

4

\sqrt{1} : \sqrt{2} : \sqrt{3}

5

1 : 4 : 9

Explanation:

C (i) Distance travelled in $1 \sec$ ,
$s_{1} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times (1)^{2}$
$s_{1} = 5 m$
(ii) Distance travelled in $2 \sec$
$s_{2} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times (2)^{2}$ .
$s_{2} = 20 m$
(iii) Distance travelled in $3 \sec$
$s_{3} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times (3)^{2}$
$s_{3} = 45 m$
The ratio of distance,
$s_{1} : s_{2} : s_{3} = 5 : (20 - 5) : (45 - 20)$
$= 5 : 15 : 25$
$= 1 : 3 : 5$

Kerala CEE- 2014

Motion in One Dimensions

141788 A stone is thrown vertically upwards. When the stone is at a height equal to half of its maximum height its speed will be $10 m / s$ , then the maximum height attained by the stone is (Take $g = 10 m / s^{2}$ )

1

3 m

2

15 m

3

1 m

4

10 m

Explanation:

D When the stone is at a height equal to half of its maximum height
From, $v^{2} = u^{2} - 2 g h$
$v = 10 m / s$
$(10)^{2} = u^{2} - 2 \times 10 \times \frac{h}{2}$
$100 = u^{2} - 10 h$
Speed at the maximum height,
$v = 0$
$v^{2} = u^{2} - 2 g h$
$0 = u^{2} - 2 \times 10 \times h$
$u^{2} = 20 h \dots \dots \dots \dots . .$
From equation (i) and (ii), we get-
$100 = 20 h - 10 h$
$10 h = 100 \Rightarrow h = 10 m$

JCECE-2014

Motion in One Dimensions

141789 A ball is released from the top of a tower of height $h$ meters. It takes $T$ seconds to reach the ground. What is the position of the ball at $\frac{T}{3}$ second

1

\frac{8 h}{9}

meters from the ground

2

\frac{7 h}{9}

meters from the ground

3

\frac{h}{9}

meters from the ground

4

\frac{17 h}{18}

meters from the ground

Explanation:

A Given,
Height of tower $= h$
Time taken to reach the ground $= T$
From,
$h = ut + \frac{1}{2} {gt}^{2}$
$h = 0 + \frac{1}{2} {gT}^{2}$
Distance travelled in $T / 3$ second
$h = \frac{1}{2} g {(\frac{T}{3})}^{2} = \frac{1}{2} g \frac{T^{2}}{9}$
$= \frac{h}{9} [from eq$
$(i)]$
$∴$ Distance of ball from the ground $= h - \frac{h}{9} = \frac{8 h}{9}$

AIIMS-2012

Motion in One Dimensions

141785 A ball when thrown upward from a tower with a given velocity takes time $t_{1}$ to reach the ground. The same ball when thrown downward with same velocity from the tower takes time $t_{2}$ to reach the ground. If the ball is dropped freely from the same tower, the time taken to reach the ground will be

1

t_{1} - t_{2}

2

\frac{t_{1} + t_{2}}{2}

3

\sqrt{t_{1} \times t_{2}}

4

\sqrt{t_{1}^{2} - t_{2}^{2}}

Explanation:

C For case $I^{st}$ ,
original image
$- h = {vt}_{1} - \frac{1}{2} {gt}_{1}^{2}$
$+ \frac{h}{t_{1}} = - v + \frac{1}{2} {gt}_{1}$
For case II $^{nd}$ ,
$- h = - {vt}_{2} - \frac{1}{2} {gt}_{2}^{2}$
$\frac{h}{t_{2}} = v + \frac{1}{2} {gt}_{2} \dots .$
Adding equation (i) and equation (ii), we get
$\frac{h}{t_{1}} + \frac{h}{t_{2}} = \frac{1}{2} g t_{1} + \frac{1}{2} g t_{2}$
$h (\frac{1}{t_{1}} + \frac{1}{t_{2}}) = \frac{g}{2} (t_{1} + t_{2}) \Rightarrow h \frac{(t_{1} + t_{2})}{t_{1} \cdot t_{2}} = \frac{g}{2} (t_{1} + t_{2})$
$h = \frac{g t_{1} t_{2}}{2} \dots . . . (iii)$
For case III $^{rd}$ ,
$u = 0, t = ?$
$- h = 0 \times t - \frac{1}{2} {gt}^{2} \Rightarrow h = \frac{1}{2} {gt}^{2}$
$t^{2} = \frac{2}{g} \times h$
Putting the value of $h$ from equation (iii) in equation (iv) we get-
$t^{2} = \frac{2}{g} \times \frac{g t_{1} \cdot t_{2}}{2} \Rightarrow t = \sqrt{t_{1} \cdot t_{2}}$

SCRA-2015

Motion in One Dimensions

141786 A body dropped from top of a tower fall through $60 m$ during the last two seconds of its fall. The height of tower is : $(g = 10 m / s^{2})$

1

95 m

2

60 m

3

80 m

4

90 m

Explanation:

C Let the total time be $t$
original image
Initial velocity $(u) = 0$
After $(t - 2)$ sec, final velocity
$v = 0 + g \times (t - 2)$
$v = g (t - 2)$
Now $60 m$ height covered with velocity $v$ in $2 \sec -$
$∴ 60 = g (t - 2) \times 2 + \frac{1}{2} \times g (2)^{2}$
$60 = 20 (t - 2) + \frac{1}{2} \times 10 \times 4$
$t = 4 \sec$
Hence, total height $= \frac{1}{2} {gt}^{2}$
$= \frac{1}{2} \times 10 \times 16$
$= 80 m$

UPSEE - 2005

Motion in One Dimensions

141787 The ratio of distance traversed in successive intervals of time when a body falls freely under gravity from certain height is

1

1 : 2 : 3

2

1 : 5 : 9

3

1 : 3 : 5

4

\sqrt{1} : \sqrt{2} : \sqrt{3}

5

1 : 4 : 9

Explanation:

C (i) Distance travelled in $1 \sec$ ,
$s_{1} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times (1)^{2}$
$s_{1} = 5 m$
(ii) Distance travelled in $2 \sec$
$s_{2} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times (2)^{2}$ .
$s_{2} = 20 m$
(iii) Distance travelled in $3 \sec$
$s_{3} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times (3)^{2}$
$s_{3} = 45 m$
The ratio of distance,
$s_{1} : s_{2} : s_{3} = 5 : (20 - 5) : (45 - 20)$
$= 5 : 15 : 25$
$= 1 : 3 : 5$

Kerala CEE- 2014

Motion in One Dimensions

141788 A stone is thrown vertically upwards. When the stone is at a height equal to half of its maximum height its speed will be $10 m / s$ , then the maximum height attained by the stone is (Take $g = 10 m / s^{2}$ )

1

3 m

2

15 m

3

1 m

4

10 m

Explanation:

D When the stone is at a height equal to half of its maximum height
From, $v^{2} = u^{2} - 2 g h$
$v = 10 m / s$
$(10)^{2} = u^{2} - 2 \times 10 \times \frac{h}{2}$
$100 = u^{2} - 10 h$
Speed at the maximum height,
$v = 0$
$v^{2} = u^{2} - 2 g h$
$0 = u^{2} - 2 \times 10 \times h$
$u^{2} = 20 h \dots \dots \dots \dots . .$
From equation (i) and (ii), we get-
$100 = 20 h - 10 h$
$10 h = 100 \Rightarrow h = 10 m$

JCECE-2014

Motion in One Dimensions

141789 A ball is released from the top of a tower of height $h$ meters. It takes $T$ seconds to reach the ground. What is the position of the ball at $\frac{T}{3}$ second

1

\frac{8 h}{9}

meters from the ground

2

\frac{7 h}{9}

meters from the ground

3

\frac{h}{9}

meters from the ground

4

\frac{17 h}{18}

meters from the ground

Explanation:

A Given,
Height of tower $= h$
Time taken to reach the ground $= T$
From,
$h = ut + \frac{1}{2} {gt}^{2}$
$h = 0 + \frac{1}{2} {gT}^{2}$
Distance travelled in $T / 3$ second
$h = \frac{1}{2} g {(\frac{T}{3})}^{2} = \frac{1}{2} g \frac{T^{2}}{9}$
$= \frac{h}{9} [from eq$
$(i)]$
$∴$ Distance of ball from the ground $= h - \frac{h}{9} = \frac{8 h}{9}$

AIIMS-2012

Motion in One Dimensions

141785 A ball when thrown upward from a tower with a given velocity takes time $t_{1}$ to reach the ground. The same ball when thrown downward with same velocity from the tower takes time $t_{2}$ to reach the ground. If the ball is dropped freely from the same tower, the time taken to reach the ground will be

1

t_{1} - t_{2}

2

\frac{t_{1} + t_{2}}{2}

3

\sqrt{t_{1} \times t_{2}}

4

\sqrt{t_{1}^{2} - t_{2}^{2}}

Explanation:

C For case $I^{st}$ ,
original image
$- h = {vt}_{1} - \frac{1}{2} {gt}_{1}^{2}$
$+ \frac{h}{t_{1}} = - v + \frac{1}{2} {gt}_{1}$
For case II $^{nd}$ ,
$- h = - {vt}_{2} - \frac{1}{2} {gt}_{2}^{2}$
$\frac{h}{t_{2}} = v + \frac{1}{2} {gt}_{2} \dots .$
Adding equation (i) and equation (ii), we get
$\frac{h}{t_{1}} + \frac{h}{t_{2}} = \frac{1}{2} g t_{1} + \frac{1}{2} g t_{2}$
$h (\frac{1}{t_{1}} + \frac{1}{t_{2}}) = \frac{g}{2} (t_{1} + t_{2}) \Rightarrow h \frac{(t_{1} + t_{2})}{t_{1} \cdot t_{2}} = \frac{g}{2} (t_{1} + t_{2})$
$h = \frac{g t_{1} t_{2}}{2} \dots . . . (iii)$
For case III $^{rd}$ ,
$u = 0, t = ?$
$- h = 0 \times t - \frac{1}{2} {gt}^{2} \Rightarrow h = \frac{1}{2} {gt}^{2}$
$t^{2} = \frac{2}{g} \times h$
Putting the value of $h$ from equation (iii) in equation (iv) we get-
$t^{2} = \frac{2}{g} \times \frac{g t_{1} \cdot t_{2}}{2} \Rightarrow t = \sqrt{t_{1} \cdot t_{2}}$

SCRA-2015

Motion in One Dimensions

141786 A body dropped from top of a tower fall through $60 m$ during the last two seconds of its fall. The height of tower is : $(g = 10 m / s^{2})$

1

95 m

2

60 m

3

80 m

4

90 m

Explanation:

C Let the total time be $t$
original image
Initial velocity $(u) = 0$
After $(t - 2)$ sec, final velocity
$v = 0 + g \times (t - 2)$
$v = g (t - 2)$
Now $60 m$ height covered with velocity $v$ in $2 \sec -$
$∴ 60 = g (t - 2) \times 2 + \frac{1}{2} \times g (2)^{2}$
$60 = 20 (t - 2) + \frac{1}{2} \times 10 \times 4$
$t = 4 \sec$
Hence, total height $= \frac{1}{2} {gt}^{2}$
$= \frac{1}{2} \times 10 \times 16$
$= 80 m$

UPSEE - 2005

Motion in One Dimensions

141787 The ratio of distance traversed in successive intervals of time when a body falls freely under gravity from certain height is

1

1 : 2 : 3

2

1 : 5 : 9

3

1 : 3 : 5

4

\sqrt{1} : \sqrt{2} : \sqrt{3}

5

1 : 4 : 9

Explanation:

C (i) Distance travelled in $1 \sec$ ,
$s_{1} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times (1)^{2}$
$s_{1} = 5 m$
(ii) Distance travelled in $2 \sec$
$s_{2} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times (2)^{2}$ .
$s_{2} = 20 m$
(iii) Distance travelled in $3 \sec$
$s_{3} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times (3)^{2}$
$s_{3} = 45 m$
The ratio of distance,
$s_{1} : s_{2} : s_{3} = 5 : (20 - 5) : (45 - 20)$
$= 5 : 15 : 25$
$= 1 : 3 : 5$

Kerala CEE- 2014

Motion in One Dimensions

141788 A stone is thrown vertically upwards. When the stone is at a height equal to half of its maximum height its speed will be $10 m / s$ , then the maximum height attained by the stone is (Take $g = 10 m / s^{2}$ )

1

3 m

2

15 m

3

1 m

4

10 m

Explanation:

D When the stone is at a height equal to half of its maximum height
From, $v^{2} = u^{2} - 2 g h$
$v = 10 m / s$
$(10)^{2} = u^{2} - 2 \times 10 \times \frac{h}{2}$
$100 = u^{2} - 10 h$
Speed at the maximum height,
$v = 0$
$v^{2} = u^{2} - 2 g h$
$0 = u^{2} - 2 \times 10 \times h$
$u^{2} = 20 h \dots \dots \dots \dots . .$
From equation (i) and (ii), we get-
$100 = 20 h - 10 h$
$10 h = 100 \Rightarrow h = 10 m$

JCECE-2014

Motion in One Dimensions

141789 A ball is released from the top of a tower of height $h$ meters. It takes $T$ seconds to reach the ground. What is the position of the ball at $\frac{T}{3}$ second

1

\frac{8 h}{9}

meters from the ground

2

\frac{7 h}{9}

meters from the ground

3

\frac{h}{9}

meters from the ground

4

\frac{17 h}{18}

meters from the ground

Explanation:

A Given,
Height of tower $= h$
Time taken to reach the ground $= T$
From,
$h = ut + \frac{1}{2} {gt}^{2}$
$h = 0 + \frac{1}{2} {gT}^{2}$
Distance travelled in $T / 3$ second
$h = \frac{1}{2} g {(\frac{T}{3})}^{2} = \frac{1}{2} g \frac{T^{2}}{9}$
$= \frac{h}{9} [from eq$
$(i)]$
$∴$ Distance of ball from the ground $= h - \frac{h}{9} = \frac{8 h}{9}$

AIIMS-2012

Motion in One Dimensions

141785 A ball when thrown upward from a tower with a given velocity takes time $t_{1}$ to reach the ground. The same ball when thrown downward with same velocity from the tower takes time $t_{2}$ to reach the ground. If the ball is dropped freely from the same tower, the time taken to reach the ground will be

1

t_{1} - t_{2}

2

\frac{t_{1} + t_{2}}{2}

3

\sqrt{t_{1} \times t_{2}}

4

\sqrt{t_{1}^{2} - t_{2}^{2}}

Explanation:

C For case $I^{st}$ ,
original image
$- h = {vt}_{1} - \frac{1}{2} {gt}_{1}^{2}$
$+ \frac{h}{t_{1}} = - v + \frac{1}{2} {gt}_{1}$
For case II $^{nd}$ ,
$- h = - {vt}_{2} - \frac{1}{2} {gt}_{2}^{2}$
$\frac{h}{t_{2}} = v + \frac{1}{2} {gt}_{2} \dots .$
Adding equation (i) and equation (ii), we get
$\frac{h}{t_{1}} + \frac{h}{t_{2}} = \frac{1}{2} g t_{1} + \frac{1}{2} g t_{2}$
$h (\frac{1}{t_{1}} + \frac{1}{t_{2}}) = \frac{g}{2} (t_{1} + t_{2}) \Rightarrow h \frac{(t_{1} + t_{2})}{t_{1} \cdot t_{2}} = \frac{g}{2} (t_{1} + t_{2})$
$h = \frac{g t_{1} t_{2}}{2} \dots . . . (iii)$
For case III $^{rd}$ ,
$u = 0, t = ?$
$- h = 0 \times t - \frac{1}{2} {gt}^{2} \Rightarrow h = \frac{1}{2} {gt}^{2}$
$t^{2} = \frac{2}{g} \times h$
Putting the value of $h$ from equation (iii) in equation (iv) we get-
$t^{2} = \frac{2}{g} \times \frac{g t_{1} \cdot t_{2}}{2} \Rightarrow t = \sqrt{t_{1} \cdot t_{2}}$

SCRA-2015

Motion in One Dimensions

141786 A body dropped from top of a tower fall through $60 m$ during the last two seconds of its fall. The height of tower is : $(g = 10 m / s^{2})$

1

95 m

2

60 m

3

80 m

4

90 m

Explanation:

C Let the total time be $t$
original image
Initial velocity $(u) = 0$
After $(t - 2)$ sec, final velocity
$v = 0 + g \times (t - 2)$
$v = g (t - 2)$
Now $60 m$ height covered with velocity $v$ in $2 \sec -$
$∴ 60 = g (t - 2) \times 2 + \frac{1}{2} \times g (2)^{2}$
$60 = 20 (t - 2) + \frac{1}{2} \times 10 \times 4$
$t = 4 \sec$
Hence, total height $= \frac{1}{2} {gt}^{2}$
$= \frac{1}{2} \times 10 \times 16$
$= 80 m$

UPSEE - 2005

Motion in One Dimensions

141787 The ratio of distance traversed in successive intervals of time when a body falls freely under gravity from certain height is

1

1 : 2 : 3

2

1 : 5 : 9

3

1 : 3 : 5

4

\sqrt{1} : \sqrt{2} : \sqrt{3}

5

1 : 4 : 9

Explanation:

C (i) Distance travelled in $1 \sec$ ,
$s_{1} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times (1)^{2}$
$s_{1} = 5 m$
(ii) Distance travelled in $2 \sec$
$s_{2} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times (2)^{2}$ .
$s_{2} = 20 m$
(iii) Distance travelled in $3 \sec$
$s_{3} = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times (3)^{2}$
$s_{3} = 45 m$
The ratio of distance,
$s_{1} : s_{2} : s_{3} = 5 : (20 - 5) : (45 - 20)$
$= 5 : 15 : 25$
$= 1 : 3 : 5$

Kerala CEE- 2014

Motion in One Dimensions

141788 A stone is thrown vertically upwards. When the stone is at a height equal to half of its maximum height its speed will be $10 m / s$ , then the maximum height attained by the stone is (Take $g = 10 m / s^{2}$ )

1

3 m

2

15 m

3

1 m

4

10 m

Explanation:

D When the stone is at a height equal to half of its maximum height
From, $v^{2} = u^{2} - 2 g h$
$v = 10 m / s$
$(10)^{2} = u^{2} - 2 \times 10 \times \frac{h}{2}$
$100 = u^{2} - 10 h$
Speed at the maximum height,
$v = 0$
$v^{2} = u^{2} - 2 g h$
$0 = u^{2} - 2 \times 10 \times h$
$u^{2} = 20 h \dots \dots \dots \dots . .$
From equation (i) and (ii), we get-
$100 = 20 h - 10 h$
$10 h = 100 \Rightarrow h = 10 m$

JCECE-2014

Motion in One Dimensions

141789 A ball is released from the top of a tower of height $h$ meters. It takes $T$ seconds to reach the ground. What is the position of the ball at $\frac{T}{3}$ second

1

\frac{8 h}{9}

meters from the ground

2

\frac{7 h}{9}

meters from the ground

3

\frac{h}{9}

meters from the ground

4

\frac{17 h}{18}

meters from the ground

Explanation:

A Given,
Height of tower $= h$
Time taken to reach the ground $= T$
From,
$h = ut + \frac{1}{2} {gt}^{2}$
$h = 0 + \frac{1}{2} {gT}^{2}$
Distance travelled in $T / 3$ second
$h = \frac{1}{2} g {(\frac{T}{3})}^{2} = \frac{1}{2} g \frac{T^{2}}{9}$
$= \frac{h}{9} [from eq$
$(i)]$
$∴$ Distance of ball from the ground $= h - \frac{h}{9} = \frac{8 h}{9}$

AIIMS-2012

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