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

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: $u_y = u \sin \theta$

Acceleration:
 Horizontal acceleration: $a_x = 0$ (no horizontal acceleration)
 Vertical acceleration: $a_y = g$ (acceleration due to gravity)

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} ]

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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