Question

what do you understand by projectile motion ?show that the trejectory of aprojectile is parabolic?​

Answer 1

Answer:

Let a body is projected with speed u m/s inclined θ with horizontal line .

Then, vertical component of u, u_yu

y

= ucosθ

Horizontal component of u ,u_xu

x

= usinθ

acceleration on horizontal, ax = 0

acceleration on vertical, ay = -g

Now, use formula ,

x = = u_xt=u

x

t

x = ucosθ.t

t = x/ucosθ------(1)

Again, y = u_yt + 1/2a_yt^2u

y

t+1/2a

y

t

2

y = usinθt - 1/2gt²

Put equation (1) here,

y = usinθ × x/ucosθ - 1/2g × x²/u²cos²θ

= tanθx - 1/2gx²/u²cos²θ

\boxed{\boxed{y=tan\theta.x-\frac{gx^2}{2u^2cos^2\theta}}}

y=tanθ.x−

2u

2

cos

2

θ

gx

2

This equation is similar to Standard equation of parabola y = ax² + bx + c her, a, b and c are constant

So, A projectile motion is parabolic motion

Explanation:

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Answer 2

$\bf\huge{\underline{\red{Answer}}}$

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. (This choice of axes is the most sensible, because acceleration due to gravity is vertical thus, there will be no acceleration along the horizontal axis when air resistance is negligible.)

Equation (5) represents the path of the projectile. If we put tan θ = A and g/2u2cos2θ = B then equation (5) can be written as y = Ax - Bx2 where A and B are constants. This is the equation of a parabola. Hence, the path of the projectile is a parabola.