Question

show that the path of a projectile is a parabola ​

Answer 1

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....

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Answer 2

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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