Motion in A Straight Line Test

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Motion in A Straight Line Test - 36

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Question 1
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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
1 / -0

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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Question 1

444 s

111 s

666 s

333 s

Solution

Question 2

16:2516 : 2516:25

7:97 : 97:9

5:75 : 75:7

25:4925 : 4925:49

Solution

Question 3

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

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

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

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

Solution

Question 4

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

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

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

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

Solution

Question 5

180180180 m

120120120 m

220220220 m

240240240 m

Solution

Question 6

32hg\displaystyle \dfrac { 3 }{ 2 } \sqrt { \dfrac { h }{ g } } 23​gh​​

gh\displaystyle\sqrt { \dfrac { g }{ h } } hg​​

gh\displaystyle \sqrt { gh } gh​

12gh\displaystyle \dfrac { 1 }{ 2 } \sqrt { gh } 21​gh​

Solution

Question 7

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

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

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

45 ms−1\displaystyle 45\ { ms }^{ -1 }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 200kmh−1\displaystyle 200{ kmh }^{ -1 }200kmh−1 on a straight road

The pilot of an aeroplane which is taking off

A cyclist negotiating a-sharp curve

Solution

Question 9

10m10 m10m

15m15 m15m

20m20 m20m

25m25 m25m

Solution

Question 10

5.80 m

12 m

20 m

18.5 m

Solution

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$4$ s

$1$ s

$6$ s

$3$ s

$16 : 25$

$7 : 9$

$5 : 7$

$25 : 49$

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

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

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

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

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

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

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

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

$180$ m

$120$ m

$220$ m

$240$ m

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

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

$\displaystyle \sqrt { gh }$

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

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

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

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

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

$10 m$

$15 m$

$20 m$

$25 m$