Question

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​

Answer 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.