Question

The equation of trajectory of an oblique projectile is y=root 3x-gx^2/2 what is the initial velocity and the angle of projection of the projectile

Answer 1

We will solve it like this:

y = xTanQ - gx² / 2u² Cos²Q

The equation is: y = √3x - gx²/2

So when we solve the equation we will get:

TanQ = √3

So Q = 60°

and u² Cos² Q = 1

u² Cos² 60 = 1

u² = 4

so u will be 2

So 2 is the initial velocity.

Answer 2

Step-by-step explanation:

y=√3x-g/2x^2

y=tanA-1/2*gx^2/u^2cos^2A

ON COMPARING THE EQUATIONS

TANA=√3

A=60

and

g/2*u^2cos^2A=g/2

u^2cos^2A=1

u^2cos^2(60)=1

u^2(1/2)^2=1

u^2*1/4=1

u^2=4

u=2

The angle of projectile is 60 and initial velocity is 2m/s.