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Four marbles are dropped from the top of a tower one after the other with an interval of one second. The first one reaches the ground after 4 seconds. When the first one reaches the ground the distances between the first and second, the second and third and the third and forth will be respectively

35, 25 and 15 m

30, 20 and 10 m

20, 10 and 5 m

40, 30 and 20 m

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A balloon rises from rest with a constant acceleration g/8. A stone is released from it when it has risen to height h. The time taken by the stone to reach the ground is

$4 \sqrt{\frac{h}{g}}$

$2 \sqrt{\frac{h}{g}}$

$\sqrt{\frac{2 h}{g}}$

$\sqrt{\frac{g}{h}}$

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Explanation

The velocity of balloon at height h, $v = \sqrt{2 (\frac{g}{8}) h}$

When the stone released from this balloon, it will go upward with velocity v = $\frac{\sqrt{g h}}{2}$ (Same as that of balloon). In this condition time taken by stone to reach the ground using 2nd equation of motion:

$- h = \frac{\sqrt{g h}}{2} t - \frac{1}{2} g t^{2} t^{2} - \sqrt{\frac{h}{g}} t - \frac{2 h}{g} = 0 t = \frac{\sqrt{\frac{h}{g}} \pm \sqrt{\frac{h}{g} + 4 \times \frac{2 h}{g}}}{2} t = \frac{\sqrt{\frac{h}{g}} \pm 3 \sqrt{\frac{h}{g}}}{2} t = \frac{\sqrt{\frac{h}{g}} + 3 \sqrt{\frac{h}{g}}}{2} = 2 \sqrt{\frac{h}{g}}$

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Two bodies are thrown simultaneously from a tower with same initial velocity v₀ : one vertically upwards, the other vertically downwards. The distance between the two bodies after time t is

$2 v_{0} t + \frac{1}{2} g t^{2}$

2v₀t

$v_{0} t + \frac{1}{2} g t^{2}$

v₀t

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Explanation

For vertically upward motion, $h_{1} = v_{0} t - \frac{1}{2} g t^{2}$ and for vertically down ward motion, $h_{2} = v_{0} t + \frac{1}{2} g t^{2}$

∴ Total distance covered in t sec $h = h_{1} + h_{2} = 2 v_{o} t$ .

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A body falls freely from the top of a tower. It covers 36% of the total height in the last second before striking the ground level. The height of the tower is

50 m

75 m

100 m

125 m

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Explanation

Let height of tower is h and body takes t time to reach to ground when it fall freely.

∴ $h = \frac{1}{2} g t^{2}$ …(i)

In last second i.e. t^{^th} sec body travels = 0.36 h

It means in rest of the time i.e. in (t – 1) sec it travels

$= h - 0 .36 h = 0 .64 h$

Now applying equation of motion for (t – 1) sec

$0 .64 h = \frac{1}{2} g {(t - 1)}^{2}$ …(ii)

From (i) and (ii) we get, t = 5 sec and h = 125 m

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A particle is projected upwards. The times corresponding to height h while ascending and while descending are t₁ and t₂ respectively. The velocity of projection will be:

gt₁

gt₂

$g (t_{1} + t_{2})$

$\frac{g (t_{1} + t_{2})}{2}$

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Explanation

$A f t e r t i m e t_{1} : T i m e t o c o m e b a c k t o h e i g h t h a f t e r t i m e t_{1} = t_{2} - t_{1} T i m e t o r e a c h h i g h e s t p o i n t f r o m h e i g h t h = t i m e t o c o m e b a c k t o h e i g h t h f r o m h i g h e s t p o i n t = \frac{t_{2} - t_{1}}{2} T i m e t o r e a c h h i g h e s t p o i n t f r o m g r o u n d = \frac{t_{2} - t_{1}}{2} + t_{1} = \frac{t_{1} + t_{2}}{2} \Rightarrow \frac{t_{1} + t_{2}}{2} = \frac{u}{g} \Rightarrow u = \frac{g {(t_{1} + t)}_{2}}{2}$

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A projectile is fired vertically upwards with an initial velocity u. After an interval of T seconds a second projectile is fired vertically upwards, also with initial velocity u.

They meet at time $t = \frac{u}{g}$ and at a height $\frac{u^{2}}{2 g} + \frac{g T^{2}}{8}$

They meet at time $t = \frac{u}{g} + \frac{T}{2}$ and at a height $\frac{u^{2}}{2 g} + \frac{g T^{2}}{8}$

They meet at time $t = \frac{u}{g} + \frac{T}{2}$ and at a height $\frac{u^{2}}{2 g} - \frac{g T^{2}}{8}$

They never meet

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Explanation

For first projectile, $h_{1} = u t - \frac{1}{2} g t^{2}$

For second projectile, $h_{2} = u (t - T) - \frac{1}{2} g {(t - T)}^{2}$

When both meet i.e. $h_{1} = h_{2}$

$u t - \frac{1}{2} g t^{2} = u (t - T) - \frac{1}{2} g {(t - T)}^{2}$

⇒ $u T + \frac{1}{2} g T^{2} = g t T$

⇒ $t = \frac{u}{g} + \frac{T}{2}$

and $h_{1} = u (\frac{u}{g} + \frac{T}{2}) - \frac{1}{2} g {(\frac{u}{2} + \frac{T}{2})}^{2}$

$= \frac{u^{2}}{2 g} - \frac{g T^{2}}{8}$

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Two cars P and Q start from a point at the same time in a straight line and their positions are represented by $x_{p} (t) = a t + b t^{2}$ and $x_{Q} (t) = f t - t^{2}$ . At what time do the cars have the same velocity?

$\frac{a - f}{1 + b}$

$\frac{a + f}{2 (b - 1)}$

$\frac{a + f}{2 (1 + b)}$

$\frac{f - a}{2 (1 + b)}$

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Explanation

(d) Velocity of each car is given by -

$v_{p} = \frac{d x (t)}{d t} = a + 2 b t$

$And, v_{Q} = \frac{{dx}_{Q} (t)}{dt} = f - 2 t$

$It is given that - v_{P} = v_{Q} \Rightarrow a + 2 bt = f - 2 t \Rightarrow t = \frac{f - a}{2 (b + 1)}$

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The motion of a particle along a straight line is described by equation

$x = 8 + 12 t - t^{3}$

where x is in metre and t is in second. The retardation of the particle when its velocity becomes zero is

24 m $s^{- 2}$

zero

6 $m s^{- 2}$

12 $m s^{- 2}$

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Explanation

Given, $x = 8 + 12 t - t^{3}$

We know $v = \frac{d x}{d t}$

So, $v = 12 - 3 t^{2}$

When v=0, t=2 sec

and $a = \frac{d v}{d t}$

and a=-6t

At t=2s,

$a = - 12 m / s^{2}$

So, retardation of the particle = 12m/ $s^{2} .$

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A particle covers half of its total distance with speed $v_{1}$ and the rest half distance with speed $v_{2} .$ Its average speed during the complete journey is

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

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

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

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

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Explanation

Velocity $v = \frac{s}{t} \Rightarrow t = \frac{s}{v}$

The average speed of the particle :

$v_{av} = \frac{s + s}{\frac{s}{v_{1}} + \frac{s}{v_{2}}} v_{av} = \frac{2 v_{1} v_{2}}{v_{1} + v_{2}}$

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A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 s is $s_{1}$ and that covered in the first 20 s is $s_{2},$ then

$s_{2} = 2 s_{1}$

$s_{2} = 3 s_{1}$

$s_{2} = 4 s_{1}$

$s_{2} = s_{1}$

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Explanation

(c)

Key Idea If the particle is moving in a straight line under the action of a constant force then

distance covered $s = ut + \frac{1}{2} {at}^{2} .$

Since the body starts from rest u=0

$∴ s = \frac{1}{2} {at}^{2} Now, s_{1} = \frac{1}{2} a {(10)}^{2} . . . . (i) and s_{2} = \frac{1}{2} a (20)^{2} . . . . (ii) Dividing Eq . (i) and Eq . (ii), we get \frac{s_{1}}{s_{2}} = \frac{(10)^{2}}{(20)^{2}} \Rightarrow s_{2} = 4 s_{1}$

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