Motion in A Straight Line Test

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Motion in A Straight Line Test - 36

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Question 1
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A balloon starts rising from the ground, vertically upwards uniformly at the rate of $$\displaystyle 1\ { ms }^{ -1 }$$ . At the end of $$4$$ seconds, a body was released from the balloon. Calculate the time taken by the released body to reach the ground. Take $$\displaystyle g = 10\ { ms }^{ -2 }$$

A
$$4$$ s

B
$$1$$ s

C
$$6$$ s

D
$$3$$ s

Solution

Height of the balloon after 4 sec $$(h)=vt = 4 m$$
And, velocity$$(u) = -1 m/s$$ (Taking downward direction as positive)
Now, using 2nd equation of motion $$s=ut+\dfrac{1}{2}t^2$$
We have,
$$h= ut + \dfrac{gt^2}{2}$$
$$\Rightarrow $$$$4= (-1)t + \dfrac{10t^2}{2}$$
$$\Rightarrow $$$$4= -t + 5t^2$$
$$\Rightarrow $$ $$ 5t^2-t-4=0$$
$$\Rightarrow (5t+4)(t-1)=0$$
$$\Rightarrow t=1 sec$$
Therefore, B is correct option.
Question 2
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Two stones A and B are dropped from the top of two different towers such that they travel $$44.1$$ m and $$63.7$$ m in the last second of their motion respectively. Find the ratio of the heights of the two towers from where the stones were dropped.

A
$$16 : 25$$

B
$$7 : 9$$

C
$$5 : 7$$

D
$$25 : 49$$

Solution

A stone is dropped from the tower, Let it takes $$t$$ seconds to reach the ground.
Using $$s=ut+\dfrac{1}{2}at^2$$,
Distance travelled in $$t$$ seconds $$\Rightarrow -S_t=0\times t-\frac{1}{2}gt^2$$ $$\Rightarrow S_t=\frac{1}{2}gt^2$$ ....1
Distance travelled in $$t-1$$ seconds $$\Rightarrow -S_{t-1}=0\times t-\frac{1}{2}g(t-1)^2$$ $$\Rightarrow S_{t-1}=\dfrac{1}{2}g(t-1)^2$$ ....2
Subtract equation 2 from 1, $$ S_t-S_{t-1}=\dfrac{1}{2}g(2t-1)$$
$$\therefore $$ The distance travelled in last seconds $$S= \dfrac{1}{2}g(2t-1)$$,
Let stone A takes time $$t_A$$ and stone B takes time $$t_B$$.
$$\Rightarrow 44.1=\dfrac{1}{2}g(2t_A-1)$$ $$\Rightarrow t_A=5$$,
$$\Rightarrow 63.7=\dfrac{1}{2}g(2t_B-1)$$ $$\Rightarrow t_B=7$$,
Using equation 1,
Distance travelled by stone A$$=S_A= \dfrac{1}{2}g\times 5^2$$
Distance travelled by stone B$$=S_B= \dfrac{1}{2}g\times 7^2$$
$$\therefore$$ $$\frac{S_A}{S_B}=\dfrac{\dfrac{1}{2}g\times 5^2}{\frac{1}{2}g\times 7^2}=\dfrac{25}{49}$$ $$\Rightarrow S_A:S_B= 25:49$$
Question 3
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An object projected vertically up from the top of the tower took $$5$$ s to reach the ground. The average velocity of the object is $$\displaystyle 5 { ms }^{ -1 })$$, find its average speed. (given $$\displaystyle g\ = 10 { ms }^{ -1 })$$.

A
$$\displaystyle 65\ { ms }^{ -1 }$$

B
$$\displaystyle 13\ { ms }^{ -1 }$$

C
$$\displaystyle 26\ { ms }^{ -1 }$$

D
$$\displaystyle 25\ { ms }^{ -1 }$$

Solution

Height of tower= displacement of an object= average velocity $$\times $$ time$$=5\times 5=25 m$$,
Let initial velocity of the object is $$u$$.
Using $$s=ut+ \dfrac{1}{2}at^2$$, $$\Rightarrow -25= u\times 5-\frac{1}{2}g\times 5^2$$ $$\Rightarrow u=20 ms^{-1}$$,
Now from the tower,Let maximum height reached by object is $$h$$.
At maximum height velocity $$v=0$$,
Using $$v^2=u^2+2as$$ $$\Rightarrow 0=20^2-2gh$$ $$\Rightarrow h=20m$$,
$$\therefore $$ Total distance $$= 2h+25= 65m$$
$$\text{Average speed}= \dfrac{\text{Total Distance}}{\text{Total Time}}$$,
$$\Rightarrow $$ Average speed$$=\dfrac{65}{5}=13ms^{-1}$$
Question 4
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A ball thrown vertically upwards with speed 'u' from the top of a tower reaches the ground in $$9$$ s. Another ball is thrown vertically downwards from the same position with speed 'u', takes $$4$$ s to reach the ground. Calculate the value of 'u' . $$\displaystyle (Take \ g= 10 { ms }^{ -1 })$$

A
$$\displaystyle 125\ { ms }^{ -1 }$$

B
$$\displaystyle 75\ { ms }^{ -1 }$$

C
$$\displaystyle 25\ { ms }^{ -1 }$$

D
$$\displaystyle 175\ { ms }^{ -1 }$$

Solution

Ball is thrown from the top of tower, $$\Rightarrow height= -h$$,
For the first case, velocity = $$u$$ , Time $$t= 9 s$$,
Using $$s= ut+\frac{1}{2}at^2$$ $$\Rightarrow -h= u\times 9- \frac{1}{2} \times 10\times 9^2$$
$$\Rightarrow -h= 9u-405$$ .....1
For the second case, velocity = $$-u$$ , Time $$t= 4 s$$,
Using $$s= ut+\frac{1}{2}at^2$$ $$\Rightarrow -h= -u\times 4- \frac{1}{2} \times 10\times 4^2$$
$$\Rightarrow -h= -4u-80$$ .....2
Equating equations 1 and 2,
$$\Rightarrow 9u-405=-4u-80$$ $$\Rightarrow u= 25 ms^{-1}$$
Question 5
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A ball which is thrown vertically up from the top of the tower reaches the ground in $$12$$ s. Another ball thrown vertically downwards from the same position with the same velocity takes $$4$$ s to reach the ground. Find the height of the tower. $$\displaystyle (Take\ g = 10 {ms}^{-2} ) $$

A
$$180$$ m

B
$$120$$ m

C
$$220$$ m

D
$$240$$ m

Solution

Ball is thrown from the top of tower, $$\Rightarrow height= -h$$,
For the first case, velocity = $$u$$ , Time $$t= 12 s$$,
Using $$s= ut+\frac{1}{2}at^2$$ $$\Rightarrow -h= u\times 12- \frac{1}{2} \times 10\times 12^2$$
$$\Rightarrow -h= 12u-720$$ .....1
For the second case, velocity = $$-u$$ , Time $$t= 4 s$$,
Using $$s= ut+\frac{1}{2}at^2$$ $$\Rightarrow -h= -u\times 4- \frac{1}{2} \times 10\times 4^2$$
$$\Rightarrow -h= -4u-80$$ .....2
Equating equations 1 and 2,
$$\Rightarrow 12u-720=-4u-80$$ $$\Rightarrow u= 40 ms^{-1}$$
Put the value of $$u=40 ms^{-1}$$ in equation 1,
$$\Rightarrow -h= 12\times 40-720$$ $$\Rightarrow h=240m$$
Question 6
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A body is dropped from a certain height 'h' meters. Assuming that the gravitational field is nullified, after the body has travelled $$\dfrac{h}{2}$$ meters such that g $$= 0, $$discuss the motion of the body. Find an expression for the time taken by the body to reach the ground.

A
$$\displaystyle \dfrac { 3 }{ 2 } \sqrt { \dfrac { h }{ g } } $$

B
$$\displaystyle\sqrt { \dfrac { g }{ h } } $$

C
$$\displaystyle \sqrt { gh } $$

D
$$\displaystyle \dfrac { 1 }{ 2 } \sqrt { gh } $$

Solution

Till $$\dfrac{h}{2}$$ meter,
$$s=\dfrac { 1 }{ 2 } g{ t }^{ 2 }$$
$$t=\sqrt { \dfrac { 2s }{ g } } =\sqrt { \dfrac { 2\times \dfrac { h }{ 2 } }{ g } } =\sqrt { \dfrac { h }{ g } } $$
so to cover $$\dfrac{h}{2}$$ meter with a=g,the time taken $${ t }_{ 1 }=\sqrt { \dfrac { h }{ g } } \\ $$
After that a=0
$$v(at\quad t=\sqrt { \dfrac { h }{ g } } )=g{ t }_{ 1 }=g\sqrt { \dfrac { h }{ g } } =\sqrt { gh } $$
with this contant speed,particle will travel next $$\dfrac{h}{2}$$ meter.
$${ t }_{ 2 }=\dfrac { \dfrac { h }{ 2 } }{ \sqrt { gh } } =\dfrac { 1 }{ 2 } \sqrt { \dfrac { h }{ g } } \\ t={ t }_{ 1 }+{ t }_{ 2 }=\sqrt { \dfrac { h }{ g } } +\dfrac { 1 }{ 2 } \sqrt { \dfrac { h }{ g } } =\dfrac { 3 }{ 2 } \sqrt { \dfrac { h }{ g } } $$
Question 7
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A ball thrown vertically upwards with speed 'u' from the top of a tower reaches the ground in $$9$$ s. Another ball is thrown vertically downwards from the same position with speed 'u'. take $$4$$ s to reach the ground. Calculate the value of 'u'. $$\displaystyle (Take \quad g = 10 { ms }^{ -2 })$$

A
$$\displaystyle 15\ { ms }^{ -1 }$$

B
$$\displaystyle 20\ { ms }^{ -1 }$$

C
$$\displaystyle 25\ { ms }^{ -1 }$$

D
$$\displaystyle 45\ { ms }^{ -1 }$$

Solution

Ball is thrown from the top of tower, $$\Rightarrow height= -h$$,
For the first case, velocity = $$u$$ , Time $$t= 9 s$$,
Using $$s= ut+\frac{1}{2}at^2$$ $$\Rightarrow -h= u\times 9- \frac{1}{2} \times 10\times 9^2$$
$$\Rightarrow -h= 9u-405$$ .....1
For the second case, velocity = $$-u$$ , Time $$t= 4 s$$,
Using $$s= ut+\frac{1}{2}at^2$$ $$\Rightarrow -h= -u\times 4- \frac{1}{2} \times 10\times 4^2$$
$$\Rightarrow -h= -4u-80$$ .....2
Equating equations 1 and 2,
$$\Rightarrow 9u-405=-4u-80$$ $$\Rightarrow u= 25 ms^{-1}$$
Question 8
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For ordinary terrestrial experiments, the observer in an inertial frame in the following cases is

A
A child revolving in a-giant wheel

B
A driver in a sports car moving with a . constant high speed of $$\displaystyle 200{ kmh }^{ -1 }$$ on a straight road

C
The pilot of an aeroplane which is taking off

D
A cyclist negotiating a-sharp curve

Solution

A and D options experience centripetal acceleration.
C experience linear acceleration
and a inertial frame n=must be non-accelerating .
B is non-accelerating hence correct answer.
Question 9
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A ball is thrown upwards. Its height varies with time as follows:
If the acceleration due to gravity is $$7.5 {m}/{{s}^{2}}$$, then the height $$h$$ is :

A
$$10 m$$

B
$$15 m$$

C
$$20 m$$

D
$$25 m$$

Solution

Velocity at highest point becomes zero

$$\therefore 0 = u- at$$

or $$u = at$$

$$= 7.5 \times 3.5 = 62.25 {m}/{s}$$
$${y}_{1} = u \times 1 - \displaystyle\frac{1}{2} \times 7.5 \times {1}^{2}$$

$${y}_{2} = u \times 2 - \displaystyle\frac{1}{2} \times 7.5 \times {2}^{2}$$

$$h = {y}_{2}- {y}_{1} = 15$$
Question 10
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A coin tossed in the air with a velocity of 30 m/s reaches back to the thrower in 0.4 sec. What is the highest point reached by the coin?

A
5.80 m

B
12 m

C
20 m

D
18.5 m

Solution

Given:
Initial velocity = $$30 m/s$$
Final velocity = $$0 m/s$$
Time = $$0.4 sec$$. (to return back)
$$\therefore$$ time for upward motion $$=\dfrac{0.4}{2} = 0.2$$
Acceleration due to gravity = $$-9.8 m/s$$$$^2$$
By using the formula,$$s=ut+\frac{1}{2} gt^2$$
$$s=30\times 0.2 +\frac{1}{2}\times -9.8 \times (0.2)^2$$
$$s=6-0.196$$
$$s=5.80 m$$
The maximum height reached by the coin is $$5.80 m$$

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Re-Attempt Weekly Quiz Competition

Motion in A Straight Line Test - 36

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Motion in A Straight Line Test - 36

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SHARING IS CARING

Question 1

$$4$$ s

$$1$$ s

$$6$$ s

$$3$$ s

Solution

Question 2

$$16 : 25$$

$$7 : 9$$

$$5 : 7$$

$$25 : 49$$

Solution

Question 3

$$\displaystyle 65\ { ms }^{ -1 }$$

$$\displaystyle 13\ { ms }^{ -1 }$$

$$\displaystyle 26\ { ms }^{ -1 }$$

$$\displaystyle 25\ { ms }^{ -1 }$$

Solution

Question 4

$$\displaystyle 125\ { ms }^{ -1 }$$

$$\displaystyle 75\ { ms }^{ -1 }$$

$$\displaystyle 25\ { ms }^{ -1 }$$

$$\displaystyle 175\ { ms }^{ -1 }$$

Solution

Question 5

$$180$$ m

$$120$$ m

$$220$$ m

$$240$$ m

Solution

Question 6

$$\displaystyle \dfrac { 3 }{ 2 } \sqrt { \dfrac { h }{ g } } $$

$$\displaystyle\sqrt { \dfrac { g }{ h } } $$

$$\displaystyle \sqrt { gh } $$

$$\displaystyle \dfrac { 1 }{ 2 } \sqrt { gh } $$

Solution

Question 7

$$\displaystyle 15\ { ms }^{ -1 }$$

$$\displaystyle 20\ { ms }^{ -1 }$$

$$\displaystyle 25\ { ms }^{ -1 }$$

$$\displaystyle 45\ { ms }^{ -1 }$$

Solution

Question 8

A child revolving in a-giant wheel

A driver in a sports car moving with a . constant high speed of $$\displaystyle 200{ kmh }^{ -1 }$$ on a straight road

The pilot of an aeroplane which is taking off

A cyclist negotiating a-sharp curve

Solution

Question 9

$$10 m$$

$$15 m$$

$$20 m$$

$$25 m$$

Solution

Question 10

5.80 m

12 m

20 m

18.5 m

Solution

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