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MOTION IN A STRIGHT LINE

269676 The average velocity of a freely falling body is numerically equal to half of the acceleration dueto gravity. The velocity of the body as it reaches the ground is

1

g

2

\frac{g}{2}

3

\frac{g}{\sqrt{2}}

4

\sqrt{2} g

MOTION IN A STRIGHT LINE

269714 A ballisthrown straight upward with a speed $v$ from a point $h$ meter abovetheground. The timetaken for theball to strike the ground is

1

\frac{v}{g} [1 + \sqrt{1 + \frac{2 h g}{V^{2}}}]

2

\frac{v}{g} [1 - \sqrt{1 - \frac{2 hg}{v^{2}}}]

3

\frac{v}{g} [1 - \sqrt{1 + \frac{2 h g}{v^{2}}}]

4

\frac{v}{g} [2 + \frac{2 h g}{v^{2}}]

Explanation:

$h = - vt + \frac{1}{2} g {tt}^{2}; g t^{2} - 2 vt - 2 h = 0$

MOTION IN A STRIGHT LINE

269715 A ballisdropped on the floor from a height of $10 m$ . It rebounds to a height of $2.5 m$ . If the ball is in contact with the floor for $0.01 s$ , then the average acceleration during contact is nearly

1

500 \sqrt{2} m / s^{2}

upwards

2

1800 \sqrt{2} m / s^{2}

dowrwards

3

1500 \sqrt{2} m / s^{2}

upwards

4

1500 \sqrt{2} m / s^{2}

dowrwards

Explanation:

$a = \frac{\sqrt{2 g h_{2}} - (- \sqrt{2 g h_{1}})}{Δ t}$

MOTION IN A STRIGHT LINE

269716 A body falling from rest has a velocity ' $v$ ' after itfallsthrough a distance' $h$ '. Thedistance it has to fall down further, for its velocity to become double, is ..... times ' $h$ '.

1 5

2 1

3 2

4 3

Explanation:

$\cdot v^{2} = 2 g h; 4 v^{2} = 2 g x$

MOTION IN A STRIGHT LINE

269730 Two balls are projected simultaneously with the same velocity ' $u$ ' from the top of a tower, one vertically upwards and the other vertically downwards. Their respectivetimes of the journeys are $t$ and $t$ . At the time of reaching the ground, the ratio of their final velocities is

1

1 : 1

2

1 : 2

3

2 : 3

4

2 : 1

Explanation:

$v = \sqrt{u^{2} + 2 g h}$ is samefor boththebodies.

MOTION IN A STRIGHT LINE

269676 The average velocity of a freely falling body is numerically equal to half of the acceleration dueto gravity. The velocity of the body as it reaches the ground is

1

g

2

\frac{g}{2}

3

\frac{g}{\sqrt{2}}

4

\sqrt{2} g

MOTION IN A STRIGHT LINE

269714 A ballisthrown straight upward with a speed $v$ from a point $h$ meter abovetheground. The timetaken for theball to strike the ground is

1

\frac{v}{g} [1 + \sqrt{1 + \frac{2 h g}{V^{2}}}]

2

\frac{v}{g} [1 - \sqrt{1 - \frac{2 hg}{v^{2}}}]

3

\frac{v}{g} [1 - \sqrt{1 + \frac{2 h g}{v^{2}}}]

4

\frac{v}{g} [2 + \frac{2 h g}{v^{2}}]

Explanation:

$h = - vt + \frac{1}{2} g {tt}^{2}; g t^{2} - 2 vt - 2 h = 0$

MOTION IN A STRIGHT LINE

269715 A ballisdropped on the floor from a height of $10 m$ . It rebounds to a height of $2.5 m$ . If the ball is in contact with the floor for $0.01 s$ , then the average acceleration during contact is nearly

1

500 \sqrt{2} m / s^{2}

upwards

2

1800 \sqrt{2} m / s^{2}

dowrwards

3

1500 \sqrt{2} m / s^{2}

upwards

4

1500 \sqrt{2} m / s^{2}

dowrwards

Explanation:

$a = \frac{\sqrt{2 g h_{2}} - (- \sqrt{2 g h_{1}})}{Δ t}$

MOTION IN A STRIGHT LINE

269716 A body falling from rest has a velocity ' $v$ ' after itfallsthrough a distance' $h$ '. Thedistance it has to fall down further, for its velocity to become double, is ..... times ' $h$ '.

1 5

2 1

3 2

4 3

Explanation:

$\cdot v^{2} = 2 g h; 4 v^{2} = 2 g x$

MOTION IN A STRIGHT LINE

269730 Two balls are projected simultaneously with the same velocity ' $u$ ' from the top of a tower, one vertically upwards and the other vertically downwards. Their respectivetimes of the journeys are $t$ and $t$ . At the time of reaching the ground, the ratio of their final velocities is

1

1 : 1

2

1 : 2

3

2 : 3

4

2 : 1

Explanation:

$v = \sqrt{u^{2} + 2 g h}$ is samefor boththebodies.

MOTION IN A STRIGHT LINE

269676 The average velocity of a freely falling body is numerically equal to half of the acceleration dueto gravity. The velocity of the body as it reaches the ground is

1

g

2

\frac{g}{2}

3

\frac{g}{\sqrt{2}}

4

\sqrt{2} g

MOTION IN A STRIGHT LINE

269714 A ballisthrown straight upward with a speed $v$ from a point $h$ meter abovetheground. The timetaken for theball to strike the ground is

1

\frac{v}{g} [1 + \sqrt{1 + \frac{2 h g}{V^{2}}}]

2

\frac{v}{g} [1 - \sqrt{1 - \frac{2 hg}{v^{2}}}]

3

\frac{v}{g} [1 - \sqrt{1 + \frac{2 h g}{v^{2}}}]

4

\frac{v}{g} [2 + \frac{2 h g}{v^{2}}]

Explanation:

$h = - vt + \frac{1}{2} g {tt}^{2}; g t^{2} - 2 vt - 2 h = 0$

MOTION IN A STRIGHT LINE

269715 A ballisdropped on the floor from a height of $10 m$ . It rebounds to a height of $2.5 m$ . If the ball is in contact with the floor for $0.01 s$ , then the average acceleration during contact is nearly

1

500 \sqrt{2} m / s^{2}

upwards

2

1800 \sqrt{2} m / s^{2}

dowrwards

3

1500 \sqrt{2} m / s^{2}

upwards

4

1500 \sqrt{2} m / s^{2}

dowrwards

Explanation:

$a = \frac{\sqrt{2 g h_{2}} - (- \sqrt{2 g h_{1}})}{Δ t}$

MOTION IN A STRIGHT LINE

269716 A body falling from rest has a velocity ' $v$ ' after itfallsthrough a distance' $h$ '. Thedistance it has to fall down further, for its velocity to become double, is ..... times ' $h$ '.

1 5

2 1

3 2

4 3

Explanation:

$\cdot v^{2} = 2 g h; 4 v^{2} = 2 g x$

MOTION IN A STRIGHT LINE

269730 Two balls are projected simultaneously with the same velocity ' $u$ ' from the top of a tower, one vertically upwards and the other vertically downwards. Their respectivetimes of the journeys are $t$ and $t$ . At the time of reaching the ground, the ratio of their final velocities is

1

1 : 1

2

1 : 2

3

2 : 3

4

2 : 1

Explanation:

$v = \sqrt{u^{2} + 2 g h}$ is samefor boththebodies.

MOTION IN A STRIGHT LINE

269676 The average velocity of a freely falling body is numerically equal to half of the acceleration dueto gravity. The velocity of the body as it reaches the ground is

1

g

2

\frac{g}{2}

3

\frac{g}{\sqrt{2}}

4

\sqrt{2} g

MOTION IN A STRIGHT LINE

269714 A ballisthrown straight upward with a speed $v$ from a point $h$ meter abovetheground. The timetaken for theball to strike the ground is

1

\frac{v}{g} [1 + \sqrt{1 + \frac{2 h g}{V^{2}}}]

2

\frac{v}{g} [1 - \sqrt{1 - \frac{2 hg}{v^{2}}}]

3

\frac{v}{g} [1 - \sqrt{1 + \frac{2 h g}{v^{2}}}]

4

\frac{v}{g} [2 + \frac{2 h g}{v^{2}}]

Explanation:

$h = - vt + \frac{1}{2} g {tt}^{2}; g t^{2} - 2 vt - 2 h = 0$

MOTION IN A STRIGHT LINE

269715 A ballisdropped on the floor from a height of $10 m$ . It rebounds to a height of $2.5 m$ . If the ball is in contact with the floor for $0.01 s$ , then the average acceleration during contact is nearly

1

500 \sqrt{2} m / s^{2}

upwards

2

1800 \sqrt{2} m / s^{2}

dowrwards

3

1500 \sqrt{2} m / s^{2}

upwards

4

1500 \sqrt{2} m / s^{2}

dowrwards

Explanation:

$a = \frac{\sqrt{2 g h_{2}} - (- \sqrt{2 g h_{1}})}{Δ t}$

MOTION IN A STRIGHT LINE

269716 A body falling from rest has a velocity ' $v$ ' after itfallsthrough a distance' $h$ '. Thedistance it has to fall down further, for its velocity to become double, is ..... times ' $h$ '.

1 5

2 1

3 2

4 3

Explanation:

$\cdot v^{2} = 2 g h; 4 v^{2} = 2 g x$

MOTION IN A STRIGHT LINE

269730 Two balls are projected simultaneously with the same velocity ' $u$ ' from the top of a tower, one vertically upwards and the other vertically downwards. Their respectivetimes of the journeys are $t$ and $t$ . At the time of reaching the ground, the ratio of their final velocities is

1

1 : 1

2

1 : 2

3

2 : 3

4

2 : 1

Explanation:

$v = \sqrt{u^{2} + 2 g h}$ is samefor boththebodies.

MOTION IN A STRIGHT LINE

269676 The average velocity of a freely falling body is numerically equal to half of the acceleration dueto gravity. The velocity of the body as it reaches the ground is

1

g

2

\frac{g}{2}

3

\frac{g}{\sqrt{2}}

4

\sqrt{2} g

MOTION IN A STRIGHT LINE

269714 A ballisthrown straight upward with a speed $v$ from a point $h$ meter abovetheground. The timetaken for theball to strike the ground is

1

\frac{v}{g} [1 + \sqrt{1 + \frac{2 h g}{V^{2}}}]

2

\frac{v}{g} [1 - \sqrt{1 - \frac{2 hg}{v^{2}}}]

3

\frac{v}{g} [1 - \sqrt{1 + \frac{2 h g}{v^{2}}}]

4

\frac{v}{g} [2 + \frac{2 h g}{v^{2}}]

Explanation:

$h = - vt + \frac{1}{2} g {tt}^{2}; g t^{2} - 2 vt - 2 h = 0$

MOTION IN A STRIGHT LINE

269715 A ballisdropped on the floor from a height of $10 m$ . It rebounds to a height of $2.5 m$ . If the ball is in contact with the floor for $0.01 s$ , then the average acceleration during contact is nearly

1

500 \sqrt{2} m / s^{2}

upwards

2

1800 \sqrt{2} m / s^{2}

dowrwards

3

1500 \sqrt{2} m / s^{2}

upwards

4

1500 \sqrt{2} m / s^{2}

dowrwards

Explanation:

$a = \frac{\sqrt{2 g h_{2}} - (- \sqrt{2 g h_{1}})}{Δ t}$

MOTION IN A STRIGHT LINE

269716 A body falling from rest has a velocity ' $v$ ' after itfallsthrough a distance' $h$ '. Thedistance it has to fall down further, for its velocity to become double, is ..... times ' $h$ '.

1 5

2 1

3 2

4 3

Explanation:

$\cdot v^{2} = 2 g h; 4 v^{2} = 2 g x$

MOTION IN A STRIGHT LINE

269730 Two balls are projected simultaneously with the same velocity ' $u$ ' from the top of a tower, one vertically upwards and the other vertically downwards. Their respectivetimes of the journeys are $t$ and $t$ . At the time of reaching the ground, the ratio of their final velocities is

1

1 : 1

2

1 : 2

3

2 : 3

4

2 : 1

Explanation:

$v = \sqrt{u^{2} + 2 g h}$ is samefor boththebodies.