Trajectory of Projectile!

Classical Mechanics Level 2

Which of the following is the equation of trajectory of a ground-to-ground projectile?

Details:

  • Angle of projection = θ =\theta

  • Displacement along x − x- axis = x =x

  • Corresponding displacement along y − y- axis = y =y

  • Acceleration due to gravity = g =g

y = x cos ⁡ θ − g x 2 2 u 2 tan ⁡ 2 θ y=x\cos{\theta}-\dfrac{gx^2}{2u^2{\tan}^2{\theta}} y = x tan ⁡ θ − g x 2 2 u 2 sin ⁡ 2 θ y=x\tan{\theta}-\dfrac{gx^2}{2u^2{\sin}^2{\theta}} y = x sin ⁡ θ − g x 2 2 u 2 tan ⁡ 2 θ y=x\sin{\theta}-\dfrac{gx^2}{2u^2{\tan}^2{\theta}} y = x tan ⁡ θ − g x 2 2 u 2 cos ⁡ 2 θ y=x\tan{\theta}-\dfrac{gx^2}{2u^2{\cos}^2{\theta}} y = x cos ⁡ θ − g x 2 2 u 2 sin ⁡ 2 θ y=x\cos{\theta}-\dfrac{gx^2}{2u^2{\sin}^2{\theta}} y = x sin ⁡ θ − g x 2 2 u 2 cos ⁡ 2 θ y=x\sin{\theta}-\dfrac{gx^2}{2u^2{\cos}^2{\theta}}

Vinayak Srivastava by Vinayak Srivastava , 15, India

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2 solutions

Aryan Sanghi
Jul 13, 2020

Eliminate t t from x = u t cos ⁡ θ , y = u t sin ⁡ θ − 1 2 g t 2 x = ut\cos\theta, y = ut\sin\theta - \frac12gt^2 and you'll get

y = x tan ⁡ θ − g x 2 2 u 2 cos ⁡ 2 θ y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}

Extending the equation

y = x tan ⁡ θ − g x 2 2 u 2 cos ⁡ 2 θ y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}

y = x tan ⁡ θ ( 1 − g x 2 u 2 cos ⁡ 2 θ × tan ⁡ θ ) y = x\tan\theta\bigg(1 - \frac{gx}{2u^2\cos^2\theta × \tan\theta}\bigg)

y = x tan ⁡ θ ( 1 − g x 2 u 2 cos ⁡ θ × sin ⁡ θ ) y = x\tan\theta\bigg(1 - \frac{gx}{2u^2\cos\theta × \sin\theta}\bigg)

y = x tan ⁡ θ ( 1 − x u 2 sin ⁡ 2 θ g ) y = x\tan\theta\bigg(1 - \frac{x}{\frac{u^2\sin2\theta}{g}}\bigg)

y = x tan ⁡ θ ( 1 − x R ) where R is Range Of Projectile \color{#3D99F6}{\boxed{y = x\tan\theta\bigg(1 - \frac{x}{R}\bigg)}} \text{ where R is Range Of Projectile}

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Thanks for sharing the proof!

Vinayak Srivastava - 11 months ago

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I didn't give the proof, just extended the equation. :)

Aryan Sanghi - 11 months ago

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Yes, but I was confused of extension and where R came in, now I understand that is because 2 sin ⁡ θ cos ⁡ θ = sin ⁡ 2 θ 2\sin{\theta}\cos{\theta}=\sin{2\theta} . Thanks!

Vinayak Srivastava - 11 months ago

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@Vinayak Srivastava – Ohk. You're welcome.

Aryan Sanghi - 11 months ago

For sin,cos and tan, you may use \sin{\theta},\cos{\theta} and \tan{\theta} to get : sin ⁡ θ , cos ⁡ θ , tan ⁡ θ \sin{\theta},\cos{\theta},\tan{\theta}

Vinayak Srivastava - 11 months ago

Eliminating the parameter time from the equations of motion of a projectile :

x = u t cos ⁡ θ , y = u t sin ⁡ θ − 1 2 g t 2 x=ut\cos \theta, y=ut\sin \theta -\dfrac 12 gt^2 , we get the result easily.

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