Physics, asked by Snehamudhiraj, 9 months ago

A body of mass 2gm is projected horizontally from the top of tower of height 20m with a velocity 10m/s.The charge on the body is 2C. Electric field is applied vertically downwards and of intensity 10 to the power -2 N/C . Find the time taken by the body to touch the ground (g=10m/s2)​

Answers

Answered by Anonymous
37

Given :

  • A body of mass 2gm is projected horizontally from the top of tower.
  • Height of tower = 20m
  • Charge on the body = 2 C
  • Electric Field,\sf\:10{}^{-2} is applied vertically downward

To Find :

The time taken by the body to touch the ground

Solution :

We have,

Mass of the body = 2gm=\sf\:2\times10{}^{-3}Kg

Height of the tower =20m

Charge on the body =2C

Electric Field \sf=10{}^{-2}NC{}^{-1}

According to the question :

A body of mass 2gm is projected horizontally from the top of tower with a velocity 10m/s

Downward Velocity ,u = 0

Then ,by equation of motion

\sf\:H=ut+\dfrac{1}{2}a\times\:t^2

\sf\:H=0+\dfrac{1}{2}a\times\:t^2

\sf\:t=\sqrt{\dfrac{2H}{a}}....(1)

Here ,

  • Acceleration ,a is the net acceleration acting on the body .

From the free body diagram of the

body

Net force acting on the body due to gravity and electric field is given by :

\sf\:F_{net}=mg+qE

\sf\:ma_{net}=mg+qE

\sf\:a_{net}=g+\dfrac{qE}{m}

Now put the given values,then:

\sf\:a_{net}=10+\dfrac{2\times10{}^{-2}}{2\times10{}^{-3}}

\sf\:a_{net}=20ms{}^{-2}

Now Put a=20m/s² in equation (1)

\sf\:t=\sqrt{\dfrac{2H}{a_{net}}}

\sf\:t=\sqrt{\dfrac{2\times20}{20}}

\sf\:t=\sqrt{2}

\sf\:t=1.414sec

The time taken by the body to touch the ground is 1.414 sec.

{\red{\boxed{\large{\bold{More\:About\:Topic}}}}}

1)Kinematic equations for uniformly accelerated motion .

\bf\:v=u+at

\bf\:s=ut+\frac{1}{2}at{}^{2}

\bf\:v{}^{2}=u{}^{2}+2as

and \bf\:s_{nth}=u+\frac{a}{2}(2n-1)

2) The motion is independent of the mass of the body ,as an equation of motion mass is not involved.This is why a heavy and lighter body when released from the same height, reach the ground simultaneously and with same velocity.

Attachments:
Answered by abdulrubfaheemi
0

Answer:

Given :

A body of mass 2gm is projected horizontally from the top of tower.

Height of tower = 20m

Charge on the body = 2 C

Electric Field,\sf\:10{}^{-2}10

−2

is applied vertically downward

To Find :

The time taken by the body to touch the ground

Solution :

We have,

Mass of the body = 2gm=\sf\:2\times10{}^{-3}2×10

−3

Kg

Height of the tower =20m

Charge on the body =2C

Electric Field \sf=10{}^{-2}NC{}^{-1}=10

−2

NC

−1

According to the question :

A body of mass 2gm is projected horizontally from the top of tower with a velocity 10m/s

Downward Velocity ,u = 0

Then ,by equation of motion

\sf\:H=ut+\dfrac{1}{2}a\times\:t^2H=ut+

2

1

a×t

2

\sf\:H=0+\dfrac{1}{2}a\times\:t^2H=0+

2

1

a×t

2

\sf\:t=\sqrt{\dfrac{2H}{a}}....(1)t=

a

2H

....(1)

Here ,

Acceleration ,a is the net acceleration acting on the body .

From the free body diagram of the

body

Net force acting on the body due to gravity and electric field is given by :

\sf\:F_{net}=mg+qEF

net

=mg+qE

\sf\:ma_{net}=mg+qEma

net

=mg+qE

\sf\:a_{net}=g+\dfrac{qE}{m}a

net

=g+

m

qE

Now put the given values,then:

\sf\:a_{net}=10+\dfrac{2\times10{}^{-2}}{2\times10{}^{-3}}a

net

=10+

2×10

−3

2×10

−2

\sf\:a_{net}=20ms{}^{-2}a

net

=20ms

−2

Now Put a=20m/s² in equation (1)

\sf\:t=\sqrt{\dfrac{2H}{a_{net}}}t=

a

net

2H

\sf\:t=\sqrt{\dfrac{2\times20}{20}}t=

20

2×20

\sf\:t=\sqrt{2}t=

2

\sf\:t=1.414sect=1.414sec

The time taken by the body to touch the ground is 1.414 sec.

{\red{\boxed{\large{\bold{More\:About\:Topic}}}}}

MoreAboutTopic

1)Kinematic equations for uniformly accelerated motion .

\bf\:v=u+atv=u+at

\bf\:s=ut+\frac{1}{2}at{}^{2}s=ut+

2

1

at

2

\bf\:v{}^{2}=u{}^{2}+2asv

2

=u

2

+2as

and \bf\:s_{nth}=u+\frac{a}{2}(2n-1)s

nth

=u+

2

a

(2n−1)

2) The motion is independent of the mass of the body ,as an equation of motion mass is not involved.This is why a heavy and lighter body when released from the same height, reach the ground simultaneously and with same velocity.

Similar questions