Question

horizontal range of a projectile is maximum when it is fired an angle of 30digreee,45,,figured,60 figures,90 digree​

Answer 1

Answer:

Correct option is

C

45

o

Let Velocity is V and the angle of projection with horizontal is θ.

∴Range, R=

g

2V

2

sinθcosθ

=

g

V

2

sin2θ

⇒

dθ

dR

=

g

V

2

2cos2θ=0 ⇒2θ=90

o

⇒θ=45

o

⇒

dθ

2

d

2

R

=−

g

V

2

4sin2θ, at θ=45

o

⇒

dθ

2

d

2

R

=−

g

4V

2

<0

Hence, at θ=45

o

, Range is maximum.