Question

A particle is projected from ground with an initial speed 417 m/s at an angle 60° with horizontal. The average velocity of the particle between its point of projection and height point of trajectory is.​

Answer 1

Explanation:

Correct option is

C

2

v

1+3cos

2

θ

For highest point-

T=(total−time)/2=

g

vsinθ

x=vcosθ×T=

g

v

2

sinθcosθ

y=

2g

v

2

sin

2

θ

displacement s=

x

2

+y

2

=

g

v

2

sin

2

θcos

2

θ+

4

sin

4

θ

v

av

=

T

s

=v

cos

2

θ+

4

sin

2

θ

⟹v

av

=

2

v

3cos

2

θ+1

Answer-(C)