Question

Projectile Motion , Help me in getting it's answer wih right explanation ! ​

Answer 1

question number 8= C part

question no 9=B part

question no10=D part

very sure about answer 100% correct

Answer 2

Given : Projectile motion

To Find : Choose correct option

Solution:

Let say  u is the velocity and θ is the angle

Horizontal Velocity = ucosθ

Vertical velocity = usinθ

Let say Time to flight = 2T

then time to reach max height = T

0  = usinθ - gT  

=>  T =  usinθ/g

Horizontal range = ucosθ * 2T

= ucosθ * 2usinθ/g

= u²sin2θ/g      (∵ sin2x=2sinxcosx)

Q.9)  Correct option is B) u²sin2θ/g

Horizontal range for projectile is   u²sin2θ/g

Horizontal range for projectile is  proportional to square of its velocity

Q.8  Correct option is B) Square of its velocity

Vertical Height

S = ut + (1/2)at²

=  usinθ  usinθ/g + (1/2)(-g) (  usinθ/g)²

= u²sin²θ/g   - u²sin²θ/2g

= u²sin²θ/2g

Q 10. )  Correct option is D)  u²sin²θ/2g

Learn More:

A particle is projected in the upword direction under gravity time of ...

brainly.in/question/11383332 11.

A particle is projected vertically upward and movesfreely under ... brainly.in/question/12773764