Motion in A Straight Line Test

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Motion in A Straight Line Test - 21

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Question 1
1 / -0

A balloon starts rising from the ground with an acceleration of $1.25 \;ms^{-2}$ , After $8$ seconds, a stone is released from the balloon, The stone will
$(g=10\;ms^{-2})$

A
cover a distance of 40 m in reaching the ground.

B
have a displaceent of 50 m

C
reach the ground in 4 second

D
begin to move down after being released

Solution

$s = \dfrac {1}{2} at^2$
$a = 1.25\ m/s^2 \Rightarrow s = \dfrac {1}{2} \times 1.25 \times 8^2 = 40\ m$

After $8\ sec$ speed of balloon:
$v = u - gt$
$v = -(1.25) \times 8 = - 10\ m/s$

Now, when stone is released distance to cover is $40\ m$ .
$v^2 = u^2 - 2gs$
$v^2 = (100)^2 + 2 \times 10 \times 40 \Rightarrow v = 30 m/s$
$v = u - gt \Rightarrow - 30 = +10 - 10t$
$t = 4\ s$
Question 2
1 / -0

A ball is dropped from the top of a building. The ball takes $0.5$ s to fall past the $3$ m length of a window at certain distance from the top of the building.
( $g=10\;m/s^{2}$ )
How far is the top of the window from the point at which the ball was dropped ?

A
$0.5$ m

B
$0.5225$ m

C
$0.6125$ m

D
$0.8$ m

Solution

$s = ut + \dfrac {1}{2} gt^2$
$3 = u (0.5) + \dfrac {1}{2} \times 10 (0.5)^2$
$3 = (0.5)u + 5(0.25)$
$\dfrac {1.75}{0.5} = u$
$\Rightarrow u = 3.5\ m/s$
$u^2 = u_1^2 + 2gs$
$\dfrac {(3.5)^2}{2g} = s$
$\Rightarrow s = \dfrac {3.5 \times 3.5}{2g}$ $\Rightarrow \ s = 0.6125\ m$
Question 3
1 / -0

A stone is dropped from the $16^{th}$ storey of a multistoried building reaches the ground in $4$ seconds. The no. of storeys travelled by the stone in $2^{nd}$ second is

A
4

B
3

C
5

D
2

Solution

Let height of each storey $= x$
$16x = ut + \dfrac {1}{2} gt^2 = 0 + \dfrac{1}{2} gt^2$
As $t=4 s$ ,
$16x = \dfrac{16g}{2}$
$x=\dfrac{g}{2}$

In $2^{nd}$ second,
$S_2 - S_1 = \dfrac{1}{2} g( 2^2 -1^2 ) = \dfrac {3g}{2}$
So, $\dfrac {3g}{2} = 3x$ hence stone traveled 3 storeys.
Question 4
1 / -0

A ball is dropped from the top of a building. The ball takes $0.5s$ to fall past the window of $3m$ in length at certain distance from the top of the building. ( $g=10m/s^{2}$ ) How fast was the ball going as it passed the bottom of the window?

A
3.5 $ms^{-1}$

B
8.5 $ms^{-1}$

C
5 $ms^{-1}$

D
12 $ms^{-1}$

Solution

$s = ut + \dfrac {1}{2} gt^2$
$3 = u (0.5) + \dfrac {1}{2} \times 10 (0.5)^2$

$3 = (0.5) u + 5 (0.25)$

$\dfrac {1.75}{0.5} = u$

$\Rightarrow u = 3.5\ m/s$
$v = u + gt$
$v = (3.5) + 10 (0.5) = 8.5\ m/s$
Question 5
1 / -0

A body falls from a height of $200$ m. If gravitational attraction ceases after $2$ s, further time taken by it to reach the ground is ( $g=10 \;ms^{-2}$ )

A
$5\ s$

B
$9\ s$

C
$13\ s$

D
$17\ s$

Solution

Distance travelled in $2\ sec$ .
$s = ut + \dfrac {1}{2}gt^2 = \dfrac {1}{2}g(2)^2 = 2g = 20\ mt$

Remaining $= 180\ m$
$v = u + gt = 0 + 10 \times 2 = 20 m/s$

So, after this velocity will be constant (20 m/s) downwards (since no $g$ is acting)
Hence, time taken $= \dfrac {180}{20} = 9\ sec$
Question 6
1 / -0

A ball is dropped from the top of a building. The ball takes $0.5$ s to fall past the $3$ m length of a window at certain distance from the top of the building. Time of travel above the window is (Take $g=10\;m/s^{2}$ )

A
0.5 s

B
0.25 s

C
0.35 s

D
0.75 s

Solution

$s = ut + \dfrac {1}{2} gt^2$
$3 = u (0.5) + \dfrac {1}{2} \times 10 (0.5)^2$
$3 = (0.5)u + 5(0.25)$
$\dfrac {1.75}{0.5} = u$
$\Rightarrow u = 3.5\ m/s$

$v_1 = u = 3.5 m/s$
$v_1 = u_1 + gt$
$3.5 = 10 (t) \Rightarrow t = 0.35\ sec$
Question 7
1 / -0

A body thrown vertically up with a velocity $u$ reaches the maximum height $h$ after $T$ seconds.Which of the following statements is correct?

A
At a height $\frac{h}{2}$ from the ground its velocity is $\frac{u}{2}$

B
At a time T its velocity is $u$

C
At a time 2T its velocity is $-u$

D
At a time 2T its velocity is $\frac{u}{2}$

Solution

At height $h/2$ ,
$V^2 -u^2 = -2g(\dfrac{h}{2})$
$V^2 = u^2 - gh$
Since, $0=u^2-2gh$
$V^2 = u^2 - u^2/2$
$V=u/\sqrt 2$

$V = u - gt$
At maximum height, $V = 0$
At time T, its velocity is 0
$T = \dfrac {u}{g} .................... (1)$
at $2T, V = u -gt$
$V = u - g(2T) = u - g\times 2 (\dfrac {u}{g})$ (from 1)
$\Rightarrow V = -u$
Question 8
1 / -0

A juggler throws up balls at regular intervals of time. Each ball takes $2\ s$ to reach the highest position. If the first ball is in the highest position by the time the fifth one starts, then the separation between the first and the second balls is

A
1.225m

B
2.45m

C
4.9m

D
3.8m

Solution

Let the interval between each balls be $\Delta t$ ,
$\Rightarrow$ difference between $1^{st}$ and $5^{th}$ ball $= 4 \Delta t$
$\Rightarrow 4 \Delta t = 2 \Rightarrow \Delta t = \frac {1}{2} : Also, 4 \Delta t = \dfrac {u}{g}$
$u = 4 \Delta t g - (1)$

So, separation between $1^{st}$ and $2^{nd}$ ball,
$s_1 = u (2) - \dfrac {1}{2}g (2)^2$
$s_2 = u (2 - \Delta t) - \dfrac {1}{2} g (2 - \Delta t)^2$

So, $s_1 - s_2 = u (\Delta t) - \dfrac {1}{2}g (4 \Delta t - \Delta t^2)$
From (1) $= 4 (\Delta t)^2g - \dfrac {g}{2} (4 \Delta t - \Delta t^2)$
$= 4 (\dfrac {1}{2})^2 g - \dfrac {g}{2} (\dfrac {4}{2} - \dfrac {1}{4}) = g - \dfrac {g}{2} (\dfrac {7}{4})$
$= 1.225\ mt$
Question 9
1 / -0

A bag is dropped from a helicopter rising vertically at a constant speed of $2$ $m/s$ . The distance between the two after $2$ s is (Given $g=9.8m/s^2$ )

A
$4.9\ m$

B
$19.6\ m$

C
$29.4\ m$

D
$39.2\ m$

Solution

for helicopter
$a=0,\ t=2\mathrm{s}\mathrm{e}\mathrm{c},\ u=2m/s,\ h_{1}=ut$
for body
$h_{2}=-ut+\displaystyle \frac{1}{2}gt^{2}$
distances between them $=h_{1}-h_{2}$

So,
Distance travelled by pocket in $2$ sec,
$s_1 = 2(2) - \dfrac {1}{2}g(2)^2 = 4 - 5(2)^2 = - 15.6\ m$

Distance travelled by helicopter in $2$ sec
$s_2 = 2(2) = 4\ m$

Hence, distance between the two $\Rightarrow s_2 - s_1 = 4 - (-15.6)$ $= 19.6\ m$
Question 10
1 / -0

A body throws balls vertically upwards. He throws one, while the previous one is at its highest point. Maximum height reached by a ball if he throws one ball each per second at uniform speed is

A
$19.6m$

B
$9.8m$

C
$4.9m$

D
$2.45m$

Solution

Since, he throws one ball each per second, so time taken by ball to reach the maximum height $= 1 \ sec$
$\Rightarrow h = \dfrac {u^2}{2g}$

$\Rightarrow h = ut - \dfrac {1}{2}gt^2$

At, $t = 1\ sec$
$v = u - gt = u - \dfrac {g}{2} = 9 - \dfrac {g}{2}$
$t = 1\ sec \Rightarrow u = g - (1) = \dfrac {g}{2} = 4.9\ m$

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Motion in A Straight Line Test - 21

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Motion in A Straight Line Test - 21

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Question 1

cover a distance of 40 m in reaching the ground.

have a displaceent of 50 m

reach the ground in 4 second

begin to move down after being released

Solution

Question 2

0.50.50.5 m

0.52250.52250.5225 m

0.61250.61250.6125 m

0.80.80.8 m

Solution

Question 3

4

3

5

2

Solution

Question 4

3.5 ms−1ms^{-1}ms−1

8.5 ms−1ms^{-1}ms−1

5 ms−1ms^{-1}ms−1

12 ms−1ms^{-1}ms−1

Solution

Question 5

5 s5\ s5 s

9 s9\ s9 s

13 s13\ s13 s

17 s17\ s17 s

Solution

Question 6

0.5 s

0.25 s

0.35 s

0.75 s

Solution

Question 7

At a height h2\frac{h}{2}2h​ from the ground its velocity is u2\frac{u}{2}2u​

At a time T its velocity is uuu

At a time 2T its velocity is −u-u−u

At a time 2T its velocity is u2\frac{u}{2}2u​

Solution

Question 8

1.225m

2.45m

4.9m

3.8m

Solution

Question 9

4.9 m4.9\ m4.9 m

19.6 m19.6\ m19.6 m

29.4 m29.4\ m29.4 m

39.2 m39.2\ m39.2 m

Solution

Question 10

19.6m19.6m19.6m

9.8m9.8m9.8m

4.9m4.9m4.9m

2.45m2.45m2.45m

Solution

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$0.5$ m

$0.5225$ m

$0.6125$ m

$0.8$ m

3.5 $ms^{-1}$

8.5 $ms^{-1}$

5 $ms^{-1}$

12 $ms^{-1}$

$5\ s$

$9\ s$

$13\ s$

$17\ s$

At a height $\frac{h}{2}$ from the ground its velocity is $\frac{u}{2}$

At a time T its velocity is $u$

At a time 2T its velocity is $-u$

At a time 2T its velocity is $\frac{u}{2}$

$4.9\ m$

$19.6\ m$

$29.4\ m$

$39.2\ m$

$19.6m$

$9.8m$

$4.9m$

$2.45m$