Question

Show that the trajectory of an object thrown at a certain angle with the horizontal is a parabola.

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Answer

To demonstrate that the trajectory of an object thrown at an angle with the horizontal is a parabola, we can use some fundamental equations from physics.

  1. Initial Conditions:

    • Initial velocity: $u$
    • Angle of projection: $\theta$

    The initial velocity can be broken down into:

    • Horizontal component: $u_x = u \cos \theta$
    • Vertical component: $u_y = u \sin \theta$
  2. Acceleration:

    • Horizontal acceleration: $a_x = 0$ (no horizontal acceleration)
    • Vertical acceleration: $a_y = -g$ (acceleration due to gravity)
  3. Horizontal Motion:

    From the first equation of motion: [ v_x = u_x + a_x t = u \cos \theta ] The horizontal velocity is constant because $a_x = 0$.

    The horizontal displacement ($x$) at any time $t$ can be given by: [ x = u_x t \implies t = \frac{x}{u \cos \theta} ]

  4. Vertical Motion:

    Using the second equation of motion for vertical displacement ($y$): [ y = u_y t + \frac{1}{2} a_y t^2 ] Substituting $u_y = u \sin \theta$ and $a_y = -g$: [ y = u \sin \theta \cdot t - \frac{1}{2} g t^2 ] Now, substituting $t = \frac{x}{u \cos \theta}$ from the horizontal motion: [ y = u \sin \theta \cdot \frac{x}{u \cos \theta} - \frac{1}{2} g \left( \frac{x}{u \cos \theta} \right)^2 ] Simplifying further: [ y = x \tan \theta - \frac{1}{2} \frac{g x^2}{u^2 \cos^2 \theta} ] This can be written as: [ y = x \tan \theta - \frac{g}{2 u^2 \cos^2 \theta} x^2 ]

This equation is of the form $y = ax + bx^2$, which is the standard form of a parabolic equation.

Hence, the trajectory of an object thrown at a certain angle $\theta$ with the horizontal indeed forms a parabolic path.


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