Question

Show that path of projectile is a parabola

Answer 1

Explanation:

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

Then, vertical component of u, = ucosθ

Horizontal component of u , = usinθ

acceleration on horizontal, ax = 0

acceleration on vertical, ay = -g

Now, use formula ,

x =

x = ucosθ.t

t = x/ucosθ------(1)

Again, y =

y = usinθt - 1/2gt²

Put equation (1) here,

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

= tanθx - 1/2gx²/u²cos²θ

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

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