Question

Show that the path followed by the particle given angular projection is parabolic.

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

Distance in y Axisy = 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

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

Show that the path followed by the particle given angular projection is parabolic.

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Let the Velocity of the Projectile be "u".
• Let the projectile make an Angle made by the Horizontal be θ.
• Let the Horizontal Displacement be "x".
• Let the Vertical displacement be "y".

$\huge{\bold{\underline{Explanation:-}}}$

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As this is A case of Projectile motion (or) 2 - D motion.

It has Components in Horizontal & Vertical Directions.

Components:-

• $\large{\tt u_x = u cos \theta}$
• $\large{\tt u_y = u sin \theta}$
• $\large{\tt a_x = 0}$
• $\large{\tt a_y = - g}$

Here The Acceleration along the x - axis will be zero.

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Applying Second kinematic equation in x - Direction.

$\large{\boxed{\tt S_x = u_xt + \dfrac{1}{2}a_xt^2}}$

$\large{\tt \leadsto x = u cos \theta \times t+ \dfrac{1}{2} \times 0 \times t^2}$

$\large{\tt \leadsto x = u cos \theta . t + 0}$

$\large{\tt \leadsto x = u cos \theta . t}$

$\large{ \leadsto {\underline{\underline{\tt t = \dfrac{x}{cos \theta}}}} \: \tt -------(1)}$

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Applying Second kinematic equation in y - Direction.

$\large{\boxed{\tt S_y = u_yt + \dfrac{1}{2}a_yt^2}}$

$\large{\tt \leadsto x = u sin \theta \times t+ \dfrac{1}{2} \times - g \times t^2}$

$\large{\tt \leadsto x = u sin \theta . t - \dfrac{1}{2}g. t^2}$

Substituting the value of time from Equation (1).

$\large{\tt \leadsto x = u sin \theta . t - \dfrac{1}{2}g. t^2}$

$\large{\tt \leadsto x = u sin \theta \times \dfrac{x}{u cos \theta} - \dfrac{1}{2}g. \bigg[ \dfrac{x}{u cos \theta} \bigg]^2}$

$\large{\tt \leadsto x = \cancel{u} sin \theta \times \dfrac{x}{\cancel{u} cos \theta} - \dfrac{1}{2}g. \bigg[ \dfrac{x^2}{u^2 cos^2\theta} \bigg]}$

$\large{\tt \leadsto x = sin \theta \times \dfrac{x}{ cos \theta} - \dfrac{1}{2}g. \bigg[ \dfrac{x^2}{u^2 cos^2 \theta} \bigg]}$

As we Know Sinθ/cosθ= tan θ.

Substituting,

$\large{\tt \leadsto x = (tan \theta)x - \dfrac{1}{2} . \dfrac{g}{u^2 cos^2 \theta} x^2}$

This is the Equation of Trajectory of Projectile.

This equation resembles with the Equation ax - bx².

It can be written as:-

• a = tanθ
• ${\sf b = \dfrac{g}{2u^2 cos^2 \theta}}$

Which is the Equation of Parabola .

Therefore, the Trajectory of Projectile is Parabola. & The projectile follows Parabolic path.

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