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A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time (Given $g = 9 .8 m / s^{2})$

At least 0.8 m/s

Any speed less than 19.6 m/s

Only with speed 19.6 m/s

More than 19.6 m/s

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Explanation

Interval of ball throw = 2 sec.

If we want that minimum three (more than two) ball remain in air then time of flight of first ball must be greater than 4 sec.

$T > 4$ sec

$\frac{2 u}{g} > 4 s e c \Rightarrow u > 19 .6 m / s$

for u =19.6. First ball will just strike the ground(in sky)

Second ball will be at highest point (in sky)

Third ball will be at point of projection or at ground (not in sky)

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If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is

$\frac{1}{2} g t^{2}$

$u t - \frac{1}{2} g t^{2}$

$(u - g t) t$

$u t$

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Explanation

The distance covered by the ball during the last t seconds of its upward motion = Distance covered by it in first t seconds of its downward motion

From $h = u t + \frac{1}{2} g t^{2}$

$h = \frac{1}{2} g t^{2}$ [As u = 0 for it downward motion]

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Velocity-time curve for a body projected vertically upwards is

Parabola

Ellipse

Hyperbola

Straight line

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Explanation

Because acceleration due to gravity is constant so the slope of line will be constant i.e. velocity time curve for a body projected vertically upwards is straight line.

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A car travels a distance S on a straight road in two hours and then returns to the starting point in the next three hours. Its average velocity is

S/5

2S/5

S/2 + S/3

None of the above

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Explanation

Average velocity $= \frac{Total displacement}{Time} = \frac{0}{2 + 3} = 0$

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A body has speed V, 2V and 3V in first 1/3 of distance S, second 1/3 of S and third 1/3 of S respectively. Its average speed will be

$\frac{18}{11} V$

$\frac{11}{18} V$

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Explanation

$v_{a v} = \frac{Total distance}{Time taken} = \frac{x}{\frac{x / 3}{v} + \frac{x / 3}{2 v} + \frac{x / 3}{3 v}} = \frac{18}{11} v$

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If the body covers one-third distance at speed v₁, next one third at speed v₂ and last one third at speed v₃, then average speed will be

$\frac{v_{1} v_{2} + v_{2} v_{3} + v_{3} v_{1}}{v_{1} + v_{2} + v_{3}}$

$\frac{v_{1} + v_{2} + v_{3}}{3}$

$\frac{v_{1} v_{2} v_{3}}{v_{1} v_{2} + v_{2} v_{3} + v_{3} v_{1}}$

$\frac{3 v_{1} v_{2} v_{3}}{v_{1} v_{2} + v_{2} v_{3} + v_{3} v_{1}}$

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Explanation

$v_{a v} = \frac{x}{\frac{x / 3}{v_{1}} + \frac{x / 3}{v_{2}} + \frac{x / 3}{v_{3}}} = \frac{3 v_{1} v_{2} v_{3}}{v_{1} v_{2} + v_{2} v_{3} + v_{1} v_{3}}$

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The displacement of the particle varies with time according to the relation $x = \frac{k}{b} [1 - e^{- b t}]$ . Then the velocity of the particle is

$k (e^{- b t})$

$\frac{k}{b^{2} e^{- b t}}$

$k b e^{- b t}$

None of these

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Explanation

$v = \frac{d x}{d t} = \frac{d}{d t} [\frac{k}{b} (1 - e^{- b t})] = \frac{k}{b} [0 - (- b) e^{- b t}] = k e^{- b t}$ .

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The acceleration of a particle starting from rest, varies with time according to the relation A = – aω² sinω t. The displacement of this particle at a time t will be

$- \frac{1}{2} (a ω^{2} \sin ω t) t^{2}$

$a ω \sin ω t$

$a ω \cos ω t$

$a \sin ω t$

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Explanation

Velocity $v = \int A d t = \int (- a ω^{2} \sin ω t) d t = a ω \cos ω t$

Displacement x $= \int v d t = \int a ω \cos ω t d t = a \sin ω t$

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If the velocity of a particle is (10 + 2t²) m/s, then the average acceleration of the particle between 2s and 5s is

2 m/s²

4 m/s²

12 m/s²

14 m/s²

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Explanation

Average acceleration $= \frac{Change in velocity}{Time taken} = \frac{v_{2} - v_{1}}{t_{2} - t_{1}}$

$= \frac{[10 + 2 {(5)}^{2}] - [10 + 2 {(2)}^{2}]}{3} = \frac{60 - 18}{3} = 14 m / s^{2}$

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A bullet moving with a velocity of 200 cm/s penetrates a wooden block and comes to rest after traversing 4 cm inside it. What velocity is needed for travelling distance of 9 cm in same block?

100 cm/s

136.2 cm/s

300 cm/s

250 cm/s

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Explanation

As $v^{2} = u^{2} - 2 a s \Rightarrow u^{2} = 2 a s$ $(∵ v = 0)$

$\Rightarrow u^{2} \propto s$ ⇒ $\frac{u_{2}}{u_{1}} = {(\frac{s_{2}}{s_{1}})}^{1 / 2}$

⇒ $u_{2} = {(\frac{9}{4})}^{1 / 2} u_{1} = \frac{3}{2} u_{1} = 300 c m / s$

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