Question

A projectile is ptojected with a initial velocity(2i^ +3j^).Find the maximum height attained.​

Answer 1

Answer:

Velocity of projectile v=3

i

^

+10

j

^

m/s

Speed of projectile u=∣v∣=

10

2

+3

2

=

109

m/s

Angle of projectile tanθ=

3

10

⟹ sinθ=

109

10

and cosθ=

109

3

Maximum height attained H=

2g

u

2

sin

2

θ

∴ H=

2(10)

109×

109

100

=5m

Horizontal range R=

g

u

2

sin2θ

=

g

2u

2

sinθcosθ

⟹ R=

10

2(109)×

109

10

×

109

3

=6m

Explanation:

please give this answer brilliant

Answer 2

the answer is 9÷20,because maximum height of a projectile formula (vertical velocity)^2÷2×ACCELERATION of gravity