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
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, is applied vertically downward
To Find :
The time taken by the body to touch the ground
Solution :
We have,
Mass of the body = 2gm=Kg
Height of the tower =20m
Charge on the body =2C
Electric Field
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
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 :
Now put the given values,then:
Now Put a=20m/s² in equation (1)
The time taken by the body to touch the ground is 1.414 sec.
1)Kinematic equations for uniformly accelerated motion .
and
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.
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.
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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.