A particle is projected from groundwith speed 80m/s at an angle 30 dege
Answers
Explanation:
The time of flight = .
So at 4 seconds the body is at the higest point.
At t=2s and t= 6 second the body is at the same height. So Vertical displacement between 2s and 6 s is zero.( Body went up and returned to the same vertical level.
So vertical average velocity for this period is zero.
Horizontal velocity during this period is constant which is equal to .
This is the average velocity between t = 2 s and t = 6 s.
Let's see the time taken by the particle to reach maximum height,it is,
t
=
u
sin
θ
g
Given,
u
=
80
m
s
−
1
,
θ
=
30
so,
t
=
4.07
s
That means at
6
s
it already started moving down.
So,upward displacement in
2
s
is,
s
=
(
u
sin
θ
)
⋅
2
−
1
2
g
(
2
)
2
=
60.4
m
and displacement in
6
s
is
s
=
(
u
sin
θ
)
⋅
6
−
1
2
g
(
6
)
2
=
63.6
m
So,vertical dispacement in
(
6
−
2
)
=
4
s
is
(
63.6
−
60.4
)
=
3.2
m
And horizontal displacement in
(
6
−
2
)
=
4
s
is
(
u
cos
θ
⋅
4
)
=
277.13
m
So,net displacement is
4
s
is
√
3.2
2
+
277.13
2
=
277.15
m
So,average velcoity = total displacement /total time=
277.15
4
=
69.29
m
s
−
1