Question

show that projectile motion is parabolic motion​

Answer 1

Answer:

Given:

An object is thrown at an angle θ from the horizontal.

To find:

Equation of trajectory

Concept:

We will divide the projectile motion into 2 different linear motion in x and y axes.

Calculation:

Time = T

Velocity in x axis = v cos(θ)

Velocity in y axis = v sin(θ)

Distance in x Axis

=x= [v cos(θ)] × T ....(i)

Distance in y Axis

y = v sin(θ) × T - ½gT² ...........(ii)

Putting value of T in eq.(ii)

y = x × v sin(θ)/ v cos(θ) - ½g{x/vcos(θ)}²

y = x tan(θ) - gx²/2u²cos²(θ).

The derived equation is similar to a parabolic equation

y = ax - bx².

So the trajectory of a projectile is a Parabola.

Answer 2

Expression Is Given Below;-

$use \: the \: equation \\ s = ut + \frac{1}{2} at {}^{2} \\ \\ appling \: it \: on \: x \: axis \\ = > x = (u \: \cos \alpha ) t + \frac{1}{2} (0)t {}^{2} \\ = > x = u \cos\alpha \times t \\ = > t = \frac{x}{u \: cos \: \alpha }......(1) \\ \\ applying \: on \: y \: axis \\ = > y = (u \: sin \: \alpha )t + \frac{1}{2} ( - g)t {}^{2} \\ = > y = u \: sin \: \alpha \times t - \frac{1}{2} gt {}^{2} \\ \: \: \: \: \: \: using \: equation......(1) \\ = > y = u \: sin \: \alpha \times \frac{x}{u \: cos \: \alpha } - \frac{1}{2} \times g \times ( \frac{x}{u \: cos \: \alpha } ) {}^{2} \\ = > y = x \tan \alpha - \frac{1}{2} \frac{g {x}^{2} }{ {u}^{2} { \cos \alpha }^{2} } \\ \\ \\ comparing \: with \: parabola \: equation \: \\ y = 4a {x}^{2} \\ \\ \\ hence \: the \: path \: of \: projectile \: is \: parabola$

Hope It Helps.....