Question

Show that the path of a projectile is a parabola.

Answer 1

Let a body is projected with speed u m/s inclined θ with horizontal line .
Then, vertical component of u, $u_y$ = ucosθ
Horizontal component of u ,$u_x$ = usinθ
acceleration on horizontal, ax = 0
acceleration on vertical, ay = -g

Now, use formula ,
x = $= u_xt$
x = ucosθ.t
t = x/ucosθ------(1)

Again, y = $u_yt + 1/2a_yt^2$
y = usinθt - 1/2gt²
Put equation (1) here,
y = usinθ × x/ucosθ - 1/2g × x²/u²cos²θ
= tanθx - 1/2gx²/u²cos²θ
$\boxed{\boxed{y=tan\theta.x-\frac{gx^2}{2u^2cos^2\theta}}}$
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

Answer 2

Answer:

♤A body or object upwards at an angle theta other than 90° with horizontal is called projectile.

♤Let projectile is thrown with a initial velocity u at an angle theta there with horizontal.

U=uxi+yuk

Ux=ucos theta

Uy=usin theta

Neglecting air resistance

Along x direction

S=ut+1/2at^2

Distance covered by projectile = velocity time

t=X(x)/ucos theta ----------(1)

Along y direction

S=ut+1/2at^2

y=usin theta × t-1/2gt^2 ---------(2)

Substituting (1) in (2)

y= usin theta × x/ucos theta - 1/2 g (x/ucos theta)^2

y=x tan theta - (g/2u^2cos^2theta

Let tan theta =A

-g/2u^2cos^2 theta =B

Y= Ax+Bx^2--------(Parabola)

This equation represents equation of parabola.

Hence path of projectile(Trajectory) is a parabola....

Hope it is useful....

And plz mark me as Brainliest....