a shell bursts on contact with the ground and the fragment fly in all directions with a speeds upto 39.2 m/s show that a man 78.4 m away in danger for 4√2 sec
Answers
Answered by
1
Explanation:
Given, u=60m/s,R=180m
∴R=
g
u
2
sin2θ
180=
10
(60)
2
sin2θ
⇒sin2θ=
2
1
⇒θ=15
o
As there are always two directions of projections θ and (90
o
−θ) for the given range, hence two directions of projection, i.e., 15
o
and 90
o
−15
o
, i.e., 15
o
and 75
o
.
Let T
1
and T
2
be the times of flight in the two cases, where
T
1
=
g
2usin15
o
and T
2
=
g
2usin75
o
The man is in danger for a time
T
2
−T
1
=
g
2u
(sin75
o
−sin15
o
)
=
g
2u
(2cos45
o
sin30
o
)
[using identify, sinC−sinD=2cos
2
C+D
sin
2
C−D
]
=
10
2×60
×2×
2
1
×
2
1
=6
2
s.
Similar questions