Question

A particle is projected from the ground with an initial speed of 15 m/ s at an angle of 60 with horizontal.The average velocity of the particle between its point of projection and hightest point of trajectory is

Answer 1

Answer:

ANSWER

Height of projectile at its highest point is h=

2g

u

2

sin

2

θ

.

Horizontal distance covered till this point is x=

2

R

=

2g

u

2

sin2θ

=

g

u

2

sinθcosθ

.

Time taken to reach highest point is t=

2

T

=

g

usinθ

Hence, displacement is d=

h

2

+x

2

=

2g

u

2

sinθ

(

1+3cos

2

θ

).

Average velocity v=

t

d

=

2

u

1+3cos

2

θ

.

PLZ MAKE ME BRAINLIST